Chapter 8, Problem 38P

To determine

Calculate the magnetic flux density $B_{\text{2}}$ in iron and the angle $B_{\text{2}}$ makes with the interface.

Answer to Problem 38P

The magnetic flux density $B_{\text{2}}$ in iron is $2.51a_x+3.77a_y-0.0037a_z\text{m}{\text{Wb/m}}^{\text{2}}$ and the angle $B_{\text{2}}$ makes with the interface is $0.047°$.

Explanation of Solution

Calculation:

Given that, the magnetic field intensity for air is,

$H_1=10a_x+15a_y-3a_z\text{A}/\text{m}$

The magnetic flux density for air is,

$B_1={\mu }_1{H}_1$        (1)

Here,

${\mu }_1$ is the permeability of air, and

$H_1$ is the magnetic field intensity for air.

Substitute $10a_x+15a_y-3a_z$ for $H_1$ and ${\mu }_{\text{o}}$ for ${\mu }_1$ in equation (1).

$\begin{array}{c}{B}_1={\mu }_{\text{o}}\left(10a_x+15a_y-3a_z\right)\\ =10{\mu }_{\text{o}}{a}_x+15{\mu }_{\text{o}}{a}_y-3{\mu }_{\text{o}}{a}_z\end{array}$

Consider,

$B_1=\left(B_{\rho },B_{\varphi },B_z\right)$ in ${\text{Wb/m}}^{\text{2}}$.

Consider the expression for the magnetic boundary condition.

$B_{1n}=B_{2n}$        (2)

Therefore, the equation (2) becomes,

$B_{1n}=B_{2n}=-3{\mu }_{\text{o}}{a}_z\left\{\because \text{magentic}\text{flux}\text{density}\text{along}z\text{-direction}\right\}$

Thus, the tangential component of the magnetic flux density for air is,

$B_{1t}=10{\mu }_{\text{o}}{a}_x+15{\mu }_{\text{o}}{a}_y$

The tangential component of magnetic field intensity for air is.

$H_{1t}=K+H_{2t}$

Since K is the free current density which is equal to zero then,

$\begin{array}{l}{H}_{2t}=H_{1t}\\ \frac{B_{2t}}{{\mu }_2}=\frac{B_{1t}}{{\mu }_1}\left\{\because H=\frac{B}{\mu }\right\}\\ B_{2t}=\frac{{\mu }_2}{{\mu }_1}{B}_{1t}\end{array}$

Given,

The permeability of air is ${\mu }_1={\mu }_{\text{o}}$.

The permeability of iron is ${\mu }_2=200{\mu }_{\text{o}}$.

Substitute $200{\mu }_{\text{o}}$ for ${\mu }_2$, ${\mu }_{\text{o}}$ for ${\mu }_1$, and $10{\mu }_{\text{o}}{a}_x+15{\mu }_{\text{o}}{a}_y$ for $B_{1t}$ in above equation.

$\begin{array}{c}{B}_{2t}=\frac{200{\mu }_{\text{o}}}{{\mu }_{\text{o}}}\left(10{\mu }_{\text{o}}{a}_x+15{\mu }_{\text{o}}{a}_y\right)\\ =200\left(10{\mu }_{\text{o}}{a}_x+15{\mu }_{\text{o}}{a}_y\right)\\ =2000{\mu }_{\text{o}}{a}_x+3000{\mu }_{\text{o}}{a}_y\end{array}$

Consider the expression for the magnetic flux density $B_2$ in iron.

$B_2=B_{2t}+B_{2n}$        (3)

Here,

$B_{2t}$ is the tangential component of magnetic flux density for iron, and

$B_{2n}$ is the normal component of magnetic flux density for iron.

Substitute $2000{\mu }_{\text{o}}{a}_x+3000{\mu }_{\text{o}}{a}_y$ for $B_{2t}$ and $-3{\mu }_{\text{o}}{a}_z$ for $B_{2n}$ in equation (3).

$B_2=2000{\mu }_{\text{o}}{a}_x+3000{\mu }_{\text{o}}{a}_y-3{\mu }_{\text{o}}{a}_z$

Substitute $4\pi \times {10}^{-7}$ for ${\mu }_{\text{o}}$ in above equation.

$\begin{array}{c}{B}_2=2000\left(4\pi \times {10}^{-7}\right)a_x+3000\left(4\pi \times {10}^{-7}\right)a_y-3\left(4\pi \times {10}^{-7}\right)a_z\\ =2.51\times {10}^{-3}{a}_x+3.77\times {10}^{-3}{a}_y-3.77\times {10}^{-6}{a}_z\\ =\left(2.51a_x+3.77a_y-3.77\times {10}^{-3}{a}_z\right)\times {10}^{-3}{\text{Wb/m}}^{\text{2}}\\ =\left(2.51a_x+3.77a_y-0.0037a_z\right)\text{m}{\text{Wb/m}}^{\text{2}}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

The angle $B_{\text{2}}$ makes with the interface is calculated as follows,

$\begin{array}{l}\tan {\theta }_2=\left|\frac{B_{2n}}{B_{2t}}\right|\\ {\theta }_2={\tan }^{-1}\left|\frac{B_{2n}}{B_{2t}}\right|\end{array}$

Substitute $2000{\mu }_{\text{o}}{a}_x+3000{\mu }_{\text{o}}{a}_y$ for $B_{2t}$ and $-3{\mu }_{\text{o}}{a}_z$ for $B_{2n}$ in above equation.

$\begin{array}{c}{\theta }_2={\tan }^{-1}\left|\frac{-3{\mu }_{\text{o}}{a}_z}{2000{\mu }_{\text{o}}{a}_x+3000{\mu }_{\text{o}}{a}_y}\right|\\ ={\tan }^{-1}\left|\frac{-3a_z}{2000a_x+3000a_y}\right|\\ ={\tan }^{-1}\frac{3}{\sqrt{{2000}^2+{3000}^2}}\\ =0.047°\end{array}$

Conclusion:

Thus, the magnetic flux density $B_{\text{2}}$ in iron is $2.51a_x+3.77a_y-0.0037a_z\text{m}{\text{Wb/m}}^{\text{2}}$ and the angle $B_{\text{2}}$ makes with the interface is $0.047°$.