Chapter 8, Problem 39P

(a)

To determine

Find the magnetic field intensity H and magnetic flux density B for $z<0$.

Answer to Problem 39P

The magnetic field intensity H and magnetic flux density B for $z<0$ are $-5a_y\text{A}/\text{m}$ and $-6.28a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.

Explanation of Solution

Calculation:

Write the general expression to calculate the magnetic field intensity.

$H=\frac{1}{2}K\times a_n$        (1)

Here,

$K$ is the infinite sheet of current density, and

$a_n$ is the unit normal vector directed from the current sheet to the point of interest.

Refer to mentioned Figure given in the textbook.

Substitute $30a_x-40a_x$ for $K$ and $-a_z$ for $a_n$ in equation (1).

$\begin{array}{c}H=\frac{1}{2}\left(30a_x-40a_x\right)\times \left(-a_z\right)\left\{\because z<0\right\}\\ =\frac{1}{2}\left(-10a_x\right)\times \left(-a_z\right)\\ =5a_x\times a_z\\ =-5a_y\text{A}/\text{m}\left\{\because a_x\times a_z=-a_y\right\}\end{array}$

Given that, the permeability for $z<0$ is $\mu ={\mu }_{\text{o}}$.

Write the general expression to calculate the magnetic flux density.

$B=\mu H$        (2)

Here,

$\mu$ is the permeability, and

$H$ is the magnetic field intensity.

Substitute ${\mu }_{\text{o}}$ for $\mu$ and $-5a_y$ for $H$ in equation (2) to find magnetic flux density B.

$\begin{array}{c}B={\mu }_{\text{o}}\left(-5a_y\right)\\ =-5{\mu }_{\text{o}}{a}_y\end{array}$

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

$\begin{array}{c}B=-5\left(4\pi \times {10}^{-7}\right)a_y\\ =-6.28\times {10}^{-6}{a}_y\\ =-6.28a_y\mu \text{Wb}/{\text{m}}^2\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the magnetic field intensity H and magnetic flux density B for $z<0$ are $-5a_y\text{A}/\text{m}$ and $-6.28a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.

(b)

To determine

Find the magnetic field intensity H and magnetic flux density B for $0<z<2$.

Answer to Problem 39P

The magnetic field intensity H and magnetic flux density B for $0<z<2$ are $-35a_y\text{A}/\text{m}$ and $-110a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.

Explanation of Solution

Calculation:

Refer to mentioned Figure given in the textbook.

Substitute $-30a_x-40a_x$ for $K$ and $-a_z$ for $a_n$ in equation (1).

$\begin{array}{c}H=\frac{1}{2}\left(-30a_x-40a_x\right)\times \left(-a_z\right)\left\{\because 0<z<2\right\}\\ =\frac{1}{2}\left(-70a_x\right)\times \left(-a_z\right)\\ =35a_x\times a_z\\ =-35a_y\text{A}/\text{m}\left\{\because a_x\times a_z=-a_y\right\}\end{array}$

Given that, the permeability for $0<z<2$ is $\mu =2.5{\mu }_{\text{o}}$.

Substitute $2.5{\mu }_{\text{o}}$ for $\mu$ and $-35a_y$ for $H$ in equation (2) to find magnetic flux density B.

$\begin{array}{c}B=2.5{\mu }_{\text{o}}\left(-35a_y\right)\\ =-87.5{\mu }_{\text{o}}{a}_y\end{array}$

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

$\begin{array}{c}B=-87.5\left(4\pi \times {10}^{-7}\right)a_y\\ =-1.0996\times {10}^{-4}\times {10}^{-2}\times {10}^2{a}_y\\ \cong -110\times {10}^{-6}{a}_y\text{Wb}/{\text{m}}^{\text{2}}\\ \cong -110a_y\mu \text{Wb}/{\text{m}}^{\text{2}}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the magnetic field intensity H and magnetic flux density B for $0<z<2$ are $-35a_y\text{A}/\text{m}$ and $-110a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.

(c)

To determine

Find the magnetic field intensity H and magnetic flux density B for $z>2$.

Answer to Problem 39P

The magnetic field intensity H and magnetic flux density B for $z>2$ are $5a_y\text{A}/\text{m}$ and $6.283a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.

Explanation of Solution

Calculation:

Refer to mentioned Figure given in the textbook.

Substitute $-30a_x+40a_x$ for $K$ and $-a_z$ for $a_n$ in equation (1).

$\begin{array}{c}H=\frac{1}{2}\left(-30a_x+40a_x\right)\times \left(-a_z\right)\left\{\because z>2\right\}\\ =\frac{1}{2}\left(10a_x\right)\times \left(-a_z\right)\\ =-5a_x\times a_z\\ =5a_y\text{A}/\text{m}\left\{\because a_x\times a_z=-a_y\right\}\end{array}$

Given that, the permeability for $z>2$ is $\mu ={\mu }_{\text{o}}$.

Substitute ${\mu }_{\text{o}}$ for $\mu$ and $5a_y$ for $H$ in equation (2) to find magnetic flux density B.

$\begin{array}{c}B={\mu }_{\text{o}}\left(5a_y\right)\\ =5{\mu }_{\text{o}}{a}_y\end{array}$

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

$\begin{array}{c}B=5\left(4\pi \times {10}^{-7}\right)a_y\\ =6.283\times {10}^{-6}{a}_y\\ =6.283a_y\mu \text{Wb}/{\text{m}}^{\text{2}}\left\{\because 1\mu ={10}^{-6}\right\}\end{array}$

Conclusion:

Thus, the magnetic field intensity H and magnetic flux density B for $z>2$ are $5a_y\text{A}/\text{m}$ and $6.283a_y\mu \text{Wb}/{\text{m}}^{\text{2}}$ respectively.