The direction of the magnetic force exerted on wire segment a b .

Question

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Answer 1

Question

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

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Answer 2

Question

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

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Answer 3

Question

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

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Answer 4

Question

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

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Answer 5

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

Answer 6

Chapter 22, Problem 26P

(a)

To determine

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

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

Answer 7

Chapter 22, Problem 26P

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

The Fleming left hand rule stats that, the thumb points in the direction of the force, the index finger towards the direction of the magnetic field, and the middle finger in the direction of the current.

Thus, by using thus rule, if the middle finger shows in the direction of the current in segment ab, an index finger shows the direction of magnetic field then the thumb indicates in the direction of the force thus which is in the direction of the positive x axis.

Conclusion:

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

Answer 13

(b)

To determine

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

(b)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 26P

The direction of the torque associated with the force on the 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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ab (I)

Here, τ→ is the torque, F→ab is the force on the segment ab, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→ab equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment ab is in the direction of the negative z axis.

Answer 20

(c)

To determine

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

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

Answer 21

(c)

To determine

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

Answer 22

(c)

Expert Solution

Answer to Problem 26P

The direction of the magnetic force excreted on the segment is in the negative x axis.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (−0.500m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((−0.500m)j^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.500m)(1.15 T)(j^×j^)−(0.009A)(0.500m)(0.96 T)(j^×k^)=−0.432Ni^

From the above expression it can understood that the magnetic force in the negative of the unit vector i . Thus, the direction of the magnetic force excreted on the segment is in the negative x axis.

Answer 27

(d)

To determine

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

(d)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

Answer 28

(d)

To determine

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

Answer 29

(d)

Expert Solution

Answer to Problem 26P

The direction of the torque associated with the force on the 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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→cd (I)

Here, τ→ is the torque, F→cd is the force on the segment cd, and r→ is the distance from the axis.

Conclusion:

Substitute (0.500m)j^ for r→ , 0.432Ni^ for F→cd equation (I).

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

The torque related to this expression is negative of unit vector k. Thus the toque associated with the force on a segment cd is in the direction of the negative z axis.

Answer 34

(e)

To determine

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

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

Answer 35

(e)

To determine

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

Answer 36

(e)

Expert Solution

Answer to Problem 26P

The magnetic force cannot rotate in the loop.

Answer 37

(e)

Expert Solution

Answer 38

(e)

Expert Solution

Answer 39

Expert Solution

Answer 40

Explanation of Solution

As the force examined on the segment ab is along the positive x axis and the force examined is on the segment cd which is along negative x axis. The forces are equal but the direction of the force are opposite. Therefore, the force examined on the segment ab cancel the effect on the force on the segment cd.

Therefore, the magnetic force cannot rotate in the loop.

Conclusion:

Therefore, the magnetic force cannot rotate in the loop.

Answer 41

(f)

To determine

Whether the force affect the motion of the loop.

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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 42

(f)

To determine

Whether the force affect the motion of the loop.

Answer 43

(f)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

According to the expression (I) if the magnetic field, current and the length is constant then its magnetic force will be constant. Here, 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 48

(g)

To determine

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

(g)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

Answer 49

(g)

To determine

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

Answer 50

(g)

Expert Solution

Answer to Problem 26P

The magnetic force will 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

Explanation of Solution

Write the expression for the magnetic field in a wire,

F→=IL→×B→ (I)

Here, F is the force, L is the distance from the axis, I is the current of the segment cd, and B is the magnetic field.

Conclusion:

Substitute (1.15 T)j^ +(0.96T)k^ for B , 0.009A for I , and (0.300m)j for L→ in the above equation (I) to find the value of F .

F→=0.009A×(((0.300m)i^)×(1.15 T)j^ +(0.96T)k^)=(0.009A)(0.300m)(1.15 T)(i^×j^)−(0.009A)(0.500m)(0.96 T)(i^×k^)=−(0.26N)j^ +(0.16N)k^

From the above expression it can understood that the magnetic force in the negative of the unit vector j . Thus, the direction of the magnetic force excreted on the segment bc is in the direction along yz plane.

Conclusion:

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

Answer 55

(h)

To determine

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

(h)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

Answer 56

(h)

To determine

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

Answer 57

(h)

Expert Solution

Answer to Problem 26P

The direction of the torque associated with the force on the 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

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→bc (I)

Here, τ→ is the torque, F→bc is the force on the segment bc, and r→ is the distance from the axis.

Conclusion:

Substitute (0.300m)i^ for r→ , −(0.26N)j^ + (0.16N)k^ for F→bc equation (I).

τ→=(0.300m)j^×(−(0.26N)j^ + (0.16N)k^)=(0.300m)(0.26N)(j^×j^)+(0.300m)(0.16N)(j^ ×k^)=0.048 N×m i^

The torque related to this expression is positive of unit vector i^. Thus the toque associated with the force on a segment bc is in the direction of the positive x axis.

Answer 62

(i)

To determine

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

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

Answer 63

(i)

To determine

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

Answer 64

(i)

Expert Solution

Answer to Problem 26P

The direction of the torque associated segment ad is zero

Answer 65

(i)

Expert Solution

Answer 66

(i)

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

Write the expression for torque associated with the force,

τ→=r→×F→ad (I)

Here, τ→ is the torque, F→ad is the force on the segment ad, and r→ is the distance from the axis.

Conclusion:

The segment ad is turns along the x axis. This makes the distance of the segment ad from the center zero.

The zero distance from the center will makes the expression (I) zero.

Therefore, the direction of the torque associated segment ad is zero

Answer 69

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

Answer 70

(j)

To determine

Whether the loop located itself clockwise or anticlockwise.

Answer 71

(j)

Expert Solution

Answer to Problem 26P

The loop will rotate in the anticlockwise direction.

Answer 72

(j)

Expert Solution

Answer 73

(j)

Expert Solution

Answer 74

Expert Solution

Answer 75

Explanation of Solution

As the torque on the segment ab and cd is along the negative z direction and the torque on the segment ad is zero.

Thus, the net torque is in the direction along the positive z axis, thereby it rotate the rectangular loop in the anticlockwise direction.

Conclusion:

There, the loop will rotate in the anticlockwise direction.

Answer 76

(k)

To determine

The magnitude of the magnetic moment of the loop.

(k)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

Answer 77

(k)

To determine

The magnitude of the magnetic moment of the loop.

Answer 78

(k)

Expert Solution

Answer to Problem 26P

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

Answer 79

(k)

Expert Solution

Answer 80

(k)

Expert Solution

Answer 81

Expert Solution

Answer 82

Explanation of Solution

Write the formula to calculate the magnetic moment of the loop,

μ=NIlb (I)

Here, N is the number of turns, I is the current , l is the length of the loop, b is the breadth of the loop.

Conclusion:

Substitute 1 for N , 0.900A for I , 0.500m for l, and 0.300m for b in the expression (I).

μ=1(0.900A)(0.500m)(0.300m)=0.135A⋅m2

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

Answer 83

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

(l)

Expert Solution

Answer to Problem 26P

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

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

Answer 84

(l)

To determine

The angle between the magnetic moment vector and magnetic field.

Answer 85

(l)

Expert Solution

Answer to Problem 26P

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

Answer 86

(l)

Expert Solution

Answer 87

(l)

Expert Solution

Answer 88

Expert Solution

Answer 89

Explanation of Solution

It is given that the current if it is flowing in clockwise direction, by right hand thumb rule, the finger curled will point in the direction of the current, the thumb in the direction of the magnetic field, thus the direction of the magnetic moment in downwards.

Thus it is along the negative z direction and the angle between the magnetic moment and the magnetic field is,

ϕ=90°+40°=130°

Conclusion:

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

Answer 90

(m)

To determine

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

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

Answer 91

(m)

To determine

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

Answer 92

(m)

Expert Solution

Answer to Problem 26P

The torque on the loop is 0.155N⋅m .

Answer 93

(m)

Expert Solution

Answer 94

(m)

Expert Solution

Answer 95

Expert Solution

Answer 96

Explanation of Solution

Formula to calculate the torque in current carrying wire,

τ=μBsinϕ (I)

Here, τ is the torque, μ is the permeability, and B is the magnetic field

Conclusion:

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

τ=(0.135A⋅m2)(1.5T)sin(130°)=0.155 N⋅m

The torque on the loop is 0.155N⋅m .

Answer 97

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The direction of the magnetic force exerted on wire segment a b .

Concept explainers

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

Whether the force affect the motion of the loop.

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

Whether the loop located itself clockwise or anticlockwise.

Answer to Problem 26P

Explanation of Solution

The magnitude of the magnetic moment of the loop.

Answer to Problem 26P

Explanation of Solution

The angle between the magnetic moment vector and magnetic field.

Answer to Problem 26P

Explanation of Solution

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

Answer to Problem 26P

Explanation of Solution

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