Chapter 18.12, Problem 1QP

(a)

To determine

The magnitude of the magnetic field strength.

Answer to Problem 1QP

The magnitude of the magnetic field strength is $\underset{\_}{24,000\text{A-turns/m}}$.

Explanation of Solution

Given:

The length of coil, $l=0.25\text{m}$

The number of turns in the coil, $N=400\text{turns}$.

The current passed through wire, $I=15\text{A}$

Explanation:

Write the equation of magnetic field strength.

$H=\frac{NI}{l}$ (I)

Conclusion:

Substitute $400\text{turns}$ for N, 15 A for I, and 0.25 m for l in Equation (I).

$\begin{array}{l}H=\frac{\left(\text{400}\text{turns}\right)15\text{A}}{0.25\text{m}}\\ H=24,000\text{A-turns/m}\end{array}$

Thus, the magnitude of the magnetic field strength is $\underset{\_}{24,000\text{A-turns/m}}$.

(b)

To determine

The flux density B.

Answer to Problem 1QP

The flux density B is $\underset{\_}{3.0168\times {10}^{-2}\text{tesla}}$.

Explanation of Solution

Write the equation of flux density in vacuum.

$B={\mu }_{0}H$ (II)

Here, the permeability of a vacuum is ${\mu }_0$.

Conclusion:

Substitute $1.257\times {10}^{-6}\text{H/m}$ for ${\mu }_0$ and $24,000\text{A-turns/m}$ for H in Equation (II). $\begin{array}{c}B=\left(1.257\times {10}^{-6}\text{H/m}\right)24,000\text{A-turns/m}\\ =\text{3}\text{.0168}\times {\text{10}}^{-2}\text{tesla}\end{array}$

Thus, the flux density B is $\underset{\_}{3.0168\times {10}^{-2}\text{tesla}}$.

(c)

To determine

The flux density inside the bar of chromium.

Answer to Problem 1QP

The flux density inside the bar of chromium is $\underset{\_}{3.0177\times {10}^{-2}\text{tesla}}$.

Explanation of Solution

Write the equation of flux density in vacuum.

$\begin{array}{c}B={\mu }_{0}H+{\mu }_{0}M\\ ={\mu }_{0}H+\mu {\chi }_{m}H\\ ={\mu }_{0}H\left(1+{\chi }_m\right)\end{array}$ (III)

Here, the magnetic susceptibility is ${\chi }_m$.

Conclusion:

Refer to Table 20.2, the value of ${\chi }_m$ is $3.13\times {10}^{-4}$.

Substitute $1.257\times {10}^{-6}\text{H/m}$ for ${\mu }_0$, $24,000\text{A-turns/m}$ for H, and $3.13\times {10}^{-4}$ for ${\chi }_m$ in Equation (III).

$\begin{array}{c}B=\left(1.257\times {10}^{-6}\text{H/m}\right)\left(24,000\text{A-turns/m}\right)\left(1+3.13\times {10}^{-4}\right)\\ =3.0177\times {10}^{-2}\text{tesla}\end{array}$

Thus, the flux density inside the bar of chromium is $\underset{\_}{3.0177\times {10}^{-2}\text{tesla}}$.

(d)

To determine

The magnitude of magnetization.

Answer to Problem 1QP

The magnitude of magnetization is $\underset{\_}{7.51\text{A/m}}$.

Explanation of Solution

Write the magnitude of magnetization.

$M={\chi }_{m}H$ (IV)

Conclusion:

Substitute $24,000\text{A-turns/m}$ for H and $3.13\times {10}^{-4}$ for ${\chi }_m$ in Equation (IV).

$\begin{array}{c}M=\left(3.13\times {10}^{-4}\right)24,000\text{A-turns/m}\\ =7.51\text{A/m}\end{array}$

Thus, the magnitude of magnetization is $\underset{\_}{7.51\text{A/m}}$.