A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab ? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc ? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l). Figure P28.34

Question

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Answer 1

Textbook Question

Chapter 28, Problem 34P

A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).

Figure P28.34

Chapter 28, Problem 34P, A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

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Answer 2

Textbook Question

Chapter 28, Problem 34P

A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).

Figure P28.34

Chapter 28, Problem 34P, A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

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Answer 3

Textbook Question

Chapter 28, Problem 34P

A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).

Figure P28.34

Chapter 28, Problem 34P, A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

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Answer 4

Textbook Question

Chapter 28, Problem 34P

A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and lies in the xy plane as shown in Figure P28.34. A uniform magnetic field of magnitude 1.50 T is directed at an angle of 40.0° with respect to the y axis with field lines parallel to the yz plane. The loop carries a current of 0.900 A in the direction shown. (Ignore gravitation.) We wish to evaluate the torque on the current loop. (a) What is the direction of the magnetic force exerted on wire segment ab? (b) What is the direction of the torque associated with this force about an axis through the origin? (c) What is the direction of the magnetic force exerted on segment cd? (d) What is the direction of the torque associated with this force about an axis through the origin? (e) Can the forces examined in parts (a) and (c) combine to cause the Loop to rotate around the x axis? (f) Can they affect the motion of the loop in any way? Explain. (g) What is the direction of the magnetic force exerted on segment bc? (h) What is the direction of the torque associated with this force about an axis through the origin? (i) What is the torque on segment ad about an axis through the origin? (j) From the point of view of Figure P28.34, once the loop is released from rest at the position shown, will it rotate clockwise or counterclockwise around the x axis? (k) Compute the magnitude of the magnetic moment of the loop. (l) What is the angle between the magnetic moment sector and the magnetic field? (m) Compute the torque on the loop using the results to parts (k) and (l).

Figure P28.34

Chapter 28, Problem 34P, A rectangular loop of wire has dimensions 0.500 m by 0.300 m. The loop is pivoted at the x axis and

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

Answer 8

(a)

Expert Solution

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The direction of the magnetic force exerted on wire segment ab .

Answer 13

Answer to Problem 34P

The direction of the force is in the positive x-axis.

Answer 14

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the Fleming left hand rule, if a thumb, an index finger and a middle finger of a right hand is stretch in a position that they formed mutually perpendicular to each other, there the direction of thumb will indicate the direction of force, the direction of index finger will indicate the direction of magnetic field and the direction of middle finger will indicate the direction of current.

By using the Fleming left hand rule, if the middle finger shows the direction of current in segment ab , an index finger shows the direction of magnetic field then the thumb will show the direction of force that is in the direction of positive x-axis.

Conclusion:

Therefore, the direction of the force is in the positive x-axis.

Answer 15

(b)

Expert Solution

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The direction of the torque associated with the above force about an axis through the origin.

Answer 20

Answer to Problem 34P

The direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Answer 21

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fab→ (1)

Here,

τ→ is the associated torque.

Fab→ is the force on segment ab .

r→ is the distance from the axis.

Substitute (0.500 m)j^ for r→ and (0.432 N)i^ for Fab→ in the equation (1).

τ→=(0.500 m)j^×(0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment ab is in the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment ab about an axis through the origin is in the direction of negative z-axis.

Answer 22

(c)

Expert Solution

To determine

The direction of the magnetic force exerted on the segment cd .

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The direction of the magnetic force exerted on the segment cd .

Answer 27

Answer to Problem 34P

The direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Answer 28

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment cd is,

Fcd→=IL→×B→ (2)

Here,

I is the current on segment cd .

Fcd→ is the force on segment cd .

L→ is the distance from the axis.

B→ is the magnitude of magnetic field.

Substitute 0.900 A for I , (−0.500 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fcd→=(0.900 A)×{[−(0.500 m)j^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.500 m)(1.15 T)(j^×j^)−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=0−(0.900 A)(0.500 m)(0.96 T)(j^×k^)=−(0.432 N)i^

In the above expression it is shown that the magnetic force is in the negative of the unit vector i^ . Therefore, the magnetic force on segment cd is in the direction of negative x-axis.

Conclusion:

Therefore, the direction of the magnetic force exerted on the segment cd is in the negative x-axis.

Answer 29

(d)

Expert Solution

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

The direction of the torque associated with the force on the segment cd about an axis through the origin.

Answer 34

Answer to Problem 34P

The direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Answer 35

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fcd→ (3)

Here,

τ→ is the associated torque.

Fcd→ is the force on segment cd .

r→ is the distance from the axis.

Substitute (−0.500 m)j^ for r→ and (−0.432 N)i^ for Fcd→ in the equation (3).

τ→=(−0.500 m)j^×(−0.432 N)i^=(0.500 m)(0.432 N)(j^×i^)=−0.216 Nm k^

The torque associated in the above expression is in the negative of the unit vector k^ . Therefore, the torque associated with the force on segment cd is along the direction of negative z-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment cd about an axis through the origin is in the direction along the negative z-axis.

Answer 36

(e)

Expert Solution

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

To determine

Whether the force examined in part (a) and (c) combine to cause the loop to rotate around the x-axis.

Answer 41

Answer to Problem 34P

The magnetic force cannot rotate the loop.

Answer 42

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the force examined on the segment ab is along the positive x-axis and the force examined on the segment cd is along the negative x-axis. No matter, the magnitude of these forces are equal but the direction of these force are opposite. Therefore, the force examined on the segment ab cancel the effect of the force examined on the segment cd .

Thus the net force on the rectangular loop become zero.

Conclusion:

Therefore, the magnetic force cannot rotate the loop.

Answer 43

(f)

Expert Solution

To determine

Whether the above forces affect the motion of the loop.

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

Answer 44

(f)

Expert Solution

Answer 45

(f)

Expert Solution

Answer 46

Expert Solution

Answer 47

To determine

Whether the above forces affect the motion of the loop.

Answer 48

Answer to Problem 34P

The magnetic force will only rotate the loop and will not affect the motion of the loop.

Answer 49

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

According to the expression in the equation (2) if the magnetic field, current and the length of the loop is constant then its magnetic force will be constant too. In the given problem the magnetic field, the current and the length of the loop is constant, therefore, the magnetic force is constant. Hence this magnetic force will only rotate the loop and will not affect the motion of the loop.

Conclusion:

Therefore, the magnetic force will only rotate the loop and will not affect the motion of the loop.

Answer 50

(g)

Expert Solution

To determine

The direction of the magnetic force exerted on segment bc .

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Answer 51

(g)

Expert Solution

Answer 52

(g)

Expert Solution

Answer 53

Expert Solution

Answer 54

To determine

The direction of the magnetic force exerted on segment bc .

Answer 55

Answer to Problem 34P

The magnetic force on segment bc is in the direction along yz-plane.

Answer 56

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the magnetic force on the wire segment bc is,

Fbc→=IL→×B→ (4)

Here,

Fbc→ is the force on segment cd .

Substitute 0.900 A for I , (0.300 m)j^ for L→ and (1.15 T)j^+(0.96 T)k^ for B→ in the equation (2).

Fbc→=(0.900 A)×{[(0.300 m)i^]×[(1.15 T)j^+(0.96 T)k^]}=(0.900 A)(0.300 m)(1.15 T)(i^×j^)−(0.900 A)(0.500 m)(0.96 T)(i^×k^)=(0.16 N)k^−(0.26 N)(j^)=−(0.26 N)j^+(0.16 N)k^

In the above expression it is shown that the magnetic force is in the negative of the unit vector j^ and positive of the unit vector j^ . Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Conclusion:

Therefore, the magnetic force on segment bc is in the direction along yz-plane.

Answer 57

(h)

Expert Solution

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Answer 58

(h)

Expert Solution

Answer 59

(h)

Expert Solution

Answer 60

Expert Solution

Answer 61

To determine

The direction of the torque associated with the force on the segment bc about an axis through the origin.

Answer 62

Answer to Problem 34P

The direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Answer 63

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fbc→

Here,

Fbc→ is the force on segment bc .

Substitute (0.300 m)i^ for r→ and −(0.26 N)j^+(0.16 N)k^ for Fbc→ in the above expression.

τ→=(0.300 m)j^×[−(0.26 N)j^+(0.16 N)k^]=−(0.300 m)(0.26 N)(j^×j^)+(0.300 m)(0.16 N)(j^×k^)=0+(0.048 N⋅m)i^=0.048 N×mi^

The torque associated in the above expression is in the positive of the unit vector i^ . Therefore, the torque associated with the force on segment bc is along the direction of positive x-axis.

Conclusion:

Therefore, the direction of the torque associated with the force on segment bc about an axis through the origin is in the direction along the positive x-axis.

Answer 64

(i)

Expert Solution

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer to Problem 34P

The torque on the segment ad is zero.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

Answer 65

(i)

Expert Solution

Answer 66

(i)

Expert Solution

Answer 67

Expert Solution

Answer 68

To determine

The direction of the torque on the segment ad about an axis through the origin.

Answer 69

Answer to Problem 34P

The torque on the segment ad is zero.

Answer 70

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The Formula to calculate the torque associated with the force is,

τ→=r→×Fad→

The segment ad is pivoted along the x-axis. This makes the distance of the segment ad from the centre zero. The zero distance from the centre will the above expression of torque zero. Therefore the torque on the segment ad is zero.

Conclusion:

Therefore, the torque on the segment ad is zero.

Answer 71

(j)

Expert Solution

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

Answer 72

(j)

Expert Solution

Answer 73

(j)

Expert Solution

Answer 74

Expert Solution

Answer 75

To determine

Whether the loop locate itself clockwise or anticlockwise.

Answer 76

Answer to Problem 34P

The rectangular loop will rotate in the anticlockwise direction.

Answer 77

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

As the torque on the segment ab and cd is along the negative z-direction and the torque on segment ad is zero. Therefore, the net torque is in the direction along the positive z-axis, thereby rotate the rectangular loop in the anticlockwise direction.

Conclusion:

Therefore, the rectangular loop will rotate in the anticlockwise direction.

Answer 78

(k)

Expert Solution

To determine

The magnitude of the magnetic moment of the loop.

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Answer 79

(k)

Expert Solution

Answer 80

(k)

Expert Solution

Answer 81

Expert Solution

Answer 82

To determine

The magnitude of the magnetic moment of the loop.

Answer 83

Answer to Problem 34P

The magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Answer 84

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

The formula to calculate the magnetic moment of the loop is,

μ=NIlb

Here,

N is the number of turns.

Substitute 1 for N , 0.900 A for I , 0.500 m for l and 0.300 m for b in above equation.

μ=1×0.900 A×0.500 m×0.300 m=0.135 A⋅m2

Conclusion:

Therefore, the magnitude of the magnetic moment of the loop is 0.135 A⋅m2 .

Answer 85

(l)

Expert Solution

To determine

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

Answer 86

(l)

Expert Solution

Answer 87

(l)

Expert Solution

Answer 88

Expert Solution

Answer 89

To determine

The angle between the magnetic moment vector and magnetic field.

Answer 90

Answer to Problem 34P

The angle between the magnetic moment and magnetic field is 130° .

Answer 91

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

It is given that the current if lowing in clockwise direction therefore, here follow the principle of right hand thumb rule, if the finger is curl in the direction of current then the thumb will indicate the moment of magnetic field. Thus, the direction of magnetic moment is downward. That is along the negative z direction. The angle between the magnetic moment and magnetic field is,

ϕ=90°+40°=130°

Conclusion:

Therefore, the angle between the magnetic moment and magnetic field is 130° .

Answer 92

(m)

Expert Solution

To determine

The torque on the loop using the results of part (k) and (l).

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .

Formula to calculate the torque in current carrying wire is,

τ=μBsinϕ

Substitute 0.135 A⋅m2 for μ , 1.5 T for B and 130° for ϕ in the above expression.

τ=(0.135 A⋅m2)(1.5 T)sin130°=0.155 N⋅m

Conclusion:

Therefore, the torque on the loop is 0.155 N⋅m .

Answer 93

(m)

Expert Solution

Answer 94

(m)

Expert Solution

Answer 95

Expert Solution

Answer 96

To determine

The torque on the loop using the results of part (k) and (l).

Answer 97

Answer to Problem 34P

The torque on the loop is 0.155 N⋅m .

Answer 98

Explanation of Solution

Given Info: The length of the rectangular loop is 0.500 m , the breadth of the rectangular loop is 0.300 m , the magnitude of the magnetic field is 1.5 T , the current in the rectangular loop is 0.900 A , the directed angle of the magnetic field with respect to the y-axis is 40.0° .