The magnitude of the magnetic field strength.

Question

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Answer 1

Question

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

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Answer 2

Question

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

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Answer 3

Question

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

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Answer 4

Question

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

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Answer 5

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

Answer 6

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Answer 7

Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

Answer 8

(a)

Expert Solution

Answer to Problem 1QP

The magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The length of coil, l=0.25 m

The number of turns in the coil, N=400 turns .

The current passed through wire, I=15 A

Explanation:

Write the equation of magnetic field strength.

H=NIl (I)

Conclusion:

Substitute 400 turns for N, 15 A for I, and 0.25 m for l in Equation (I).

H=(400 turns)15 A0.25 mH=24,000 A-turns/m

Thus, the magnitude of the magnetic field strength is 24,000 A-turns/m_ .

Answer 13

(b)

To determine

The flux density B.

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

Answer 14

(b)

To determine

The flux density B.

Answer 15

(b)

Expert Solution

Answer to Problem 1QP

The flux density B is 3.0168×10−2 tesla_ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H (II)

Here, the permeability of a vacuum is μ0 .

Conclusion:

Substitute 1.257×10−6 H/m for μ0 and 24,000 A-turns/m for H in Equation (II). B=(1.257×10−6 H/m)24,000 A-turns/m=3.0168×10−2 tesla

Thus, the flux density B is 3.0168×10−2 tesla_ .

Answer 20

(c)

To determine

The flux density inside the bar of chromium.

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Answer 21

(c)

To determine

The flux density inside the bar of chromium.

Answer 22

(c)

Expert Solution

Answer to Problem 1QP

The flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the equation of flux density in vacuum.

B=μ0H+μ0M=μ0H+μχmH=μ0H(1+χm) (III)

Here, the magnetic susceptibility is χm .

Conclusion:

Refer to Table 20.2, the value of χm is 3.13×10−4 .

Substitute 1.257×10−6 H/m for μ0 , 24,000 A-turns/m for H, and 3.13×10−4 for χm in Equation (III).

B=(1.257×10−6 H/m)(24,000 A-turns/m)(1+3.13×10−4)=3.0177×10−2 tesla

Thus, the flux density inside the bar of chromium is 3.0177×10−2 tesla_ .

Answer 27

(d)

To determine

The magnitude of magnetization.

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

Answer 28

(d)

To determine

The magnitude of magnetization.

Answer 29

(d)

Expert Solution

Answer to Problem 1QP

The magnitude of magnetization is 7.51 A/m_ .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

Write the magnitude of magnetization.

M=χmH (IV)

Conclusion:

Substitute 24,000 A-turns/m for H and 3.13×10−4 for χm in Equation (IV).

M=(3.13×10−4)24,000 A-turns/m=7.51 A/m

Thus, the magnitude of magnetization is 7.51 A/m_ .

Answer 34

Ch. 18.12 - Prob. 1QP Ch. 18.12 - Prob. 2QP Ch. 18.12 - Prob. 3QP Ch. 18.12 - Prob. 4QP Ch. 18.12 - Prob. 5QP Ch. 18.12 - Prob. 6QP Ch. 18.12 - Prob. 7QP Ch. 18.12 - Prob. 8QP Ch. 18.12 - Prob. 9QP Ch. 18.12 - Prob. 10QP

The magnitude of the magnetic field strength.

The magnitude of the magnetic field strength.

Answer to Problem 1QP

Explanation of Solution

The flux density B.

Answer to Problem 1QP

Explanation of Solution

The flux density inside the bar of chromium.

Answer to Problem 1QP

Explanation of Solution

The magnitude of magnetization.

Answer to Problem 1QP

Explanation of Solution

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Chapter 18 Solutions