Chapter 9, Problem 37P

To determine

Find the magnetic field intensity and the value of $\alpha$.

Answer to Problem 37P

The magnetic field intensity and the value of $\alpha$ are $\underset{\_}{2.636\cos \left(12\pi x\right)\sin \left({10}^{11}t-331.2y\right)a_x+0.3\sin \left(12\pi x\right)\cos \left({10}^{11}t-331.2y\right)a_y\text{mA}/\text{m}}$ and $\underset{\_}{331.2}$, respectively.

Explanation of Solution

Calculation:

Write the generalized form of Maxwell’s third Equation.

$\begin{array}{c}\nabla \times E=-\frac{\partial B}{\partial t}\\ =-\frac{\partial \mu H}{\partial t}\left\{\because B=\mu H\right\}\\ =-\mu \frac{\partial H}{\partial t}\end{array}$

Rewrite the expression.

$\begin{array}{c}H=-\frac{1}{\mu }{\displaystyle \int \left(\nabla \times E\right)dt}\\ =-\frac{1}{{\mu }_{\text{o}}}{\displaystyle \int \left(\nabla \times E\right)dt}\left\{\because \mu ={\mu }_{\text{o}}\text{ for free space }\left(\text{air}\right)\right\}\end{array}$

Substitute $\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)a_z\text{V}/\text{m}$ for $E$.

$\begin{array}{c}H=-\frac{1}{{\mu }_{\text{o}}}{\displaystyle \int \left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& 0& \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\end{array}\right|dt}\\ =-\frac{1}{{\mu }_{\text{o}}}{\displaystyle \int \left(\begin{array}{l}\left\{\frac{\partial }{\partial y}\left[\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]\right\}{a}_x-\\ \left\{\frac{\partial }{\partial x}\left[\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]\right\}{a}_y+0a_z\end{array}\right)dt}\\ =-\frac{1}{{\mu }_{\text{o}}}{\displaystyle \int \left\{\begin{array}{l}\left[-\alpha \cos \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)\right]a_x\\ +12\pi \sin \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)a_y\end{array}\right\}dt}\\ =-\frac{1}{{10}^{11}{\mu }_{\text{o}}}\left\{\begin{array}{l}\left[-\alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]a_x\\ -12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_y\end{array}\right\}\end{array}$

Simplify the expression.

$H=\frac{1}{\left({10}^{11}\right)\left(4\pi \times {10}^{-7}\right)}\left\{\begin{array}{l}\left[\alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]a_x\\ +12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_y\end{array}\right\}\text{A}/\text{m}$

$H=\frac{1}{4\pi \times {10}^4}\left\{\begin{array}{l}\left[\alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]a_x\\ +12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_y\end{array}\right\}\text{A}/\text{m}$        (1)

Write the generalized form of Maxwell’s fourth Equation.

$\nabla \times H=J+\frac{\partial D}{\partial t}$

Rewrite expression for free space.

$\begin{array}{c}\nabla \times H=0+\frac{\partial D}{\partial t}\left\{\because J=0\right\}\\ =\frac{\partial \epsilon E}{\partial t}\left\{\because D=\epsilon E\right\}\\ ={\epsilon }_{\text{o}}\frac{\partial E}{\partial t}\left\{\because \epsilon ={\epsilon }_{\text{o}}\right\}\end{array}$

Rewrite the expression for electric field.

$E=\frac{1}{{\epsilon }_{\text{o}}}{\displaystyle \int \left(\nabla \times H\right)dt}$        (2)

Find $\nabla \times H$.

$\begin{array}{c}\nabla \times H=\frac{1}{4\pi \times {10}^4}\left|\begin{array}{ccc}{a}_x& a_y& a_z\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ \alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)& 12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)& 0\end{array}\right|\\ =\frac{1}{4\pi \times {10}^4}\left(\begin{array}{l}\left\{-\frac{\partial }{\partial z}\left[12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)\right]\right\}{a}_x\\ -\left\{\frac{\partial }{\partial z}\left[\alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]\right\}{a}_y\\ +\left\{\begin{array}{l}\frac{\partial }{\partial x}\left[12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)\right]-\\ \frac{\partial }{\partial y}\left[\alpha \cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)\right]\end{array}\right\}{a}_z\end{array}\right)\\ =\frac{1}{4\pi \times {10}^4}\left(\begin{array}{l}0a_x-0a_y\\ +\left\{\begin{array}{l}{\left(12\pi \right)}^2\cos \left({10}^{11}t-\alpha y\right)\cos \left(12\pi x\right)+\\ {\alpha }^2\cos \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)\end{array}\right\}{a}_z\end{array}\right)\\ =\frac{{\left(12\pi \right)}^2+{\alpha }^2}{4\pi \times {10}^4}\cos \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_z\end{array}$

Substitute $\frac{{\left(12\pi \right)}^2+{\alpha }^2}{4\pi \times {10}^4}\cos \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_z$ for $\nabla \times H$ in Equation (2).

$\begin{array}{c}E=\frac{1}{{\epsilon }_{\text{o}}}{\displaystyle \int \left[\frac{{\left(12\pi \right)}^2+{\alpha }^2}{4\pi \times {10}^4}\cos \left(12\pi x\right)\cos \left({10}^{11}t-\alpha y\right)a_z\right]dt}\\ =\frac{{\left(12\pi \right)}^2+{\alpha }^2}{4\pi \times {10}^4{\epsilon }_{\text{o}}\left({10}^{11}\right)}\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)a_z\\ =\frac{{\left(12\pi \right)}^2+{\alpha }^2}{\left(4\pi \times {10}^{15}\right)\left(\frac{{10}^{-9}}{36\pi }\right)}\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)a_z\left\{\because {\epsilon }_{\text{o}}=\frac{{10}^{-9}}{36\pi }\right\}\\ =\left(9\times {10}^{-6}\right)\left[{\left(12\pi \right)}^2+{\alpha }^2\right]\cos \left(12\pi x\right)\sin \left({10}^{11}t-\alpha y\right)a_z\text{V}/\text{m}\end{array}$

From the obtained electric field, the magnitude of electric field is $\left(9\times {10}^{-6}\right)\left[{\left(12\pi \right)}^2+{\alpha }^2\right]\text{V}/\text{m}$.

Equate the obtained magnitude of the electric field to the given magnitude.

$\left(9\times {10}^{-6}\right)\left[{\left(12\pi \right)}^2+{\alpha }^2\right]\text{V}/\text{m}=1\text{V}/\text{m}$

Rearrange the expression for $\alpha$.

$\begin{array}{c}\alpha =\sqrt{\frac{1}{\left(9\times {10}^{-6}\right)}-{\left(12\pi \right)}^2}\\ \cong 331.2\end{array}$

Substitute 331.2 for $\alpha$ in Equation (1).

$\begin{array}{c}H=\frac{1}{4\pi \times {10}^4}\left\{\begin{array}{l}\left[\left(331.2\right)\cos \left(12\pi x\right)\sin \left({10}^{11}t-331.2y\right)\right]a_x\\ +12\pi \sin \left(12\pi x\right)\cos \left({10}^{11}t-331.2y\right)a_y\end{array}\right\}\text{A}/\text{m}\\ =\left[\begin{array}{l}2.636\cos \left(12\pi x\right)\sin \left({10}^{11}t-331.2y\right)a_x\\ +0.3\sin \left(12\pi x\right)\cos \left({10}^{11}t-331.2y\right)a_y\end{array}\right]\text{mA}/\text{m}\end{array}$

Conclusion:

Thus, the magnetic field intensity and the value of $\alpha$ are $\underset{\_}{2.636\cos \left(12\pi x\right)\sin \left({10}^{11}t-331.2y\right)a_x+0.3\sin \left(12\pi x\right)\cos \left({10}^{11}t-331.2y\right)a_y\text{mA}/\text{m}}$ and $\underset{\_}{331.2}$, respectively.