Chapter 9, Problem 9.12P

To determine

(a)

The displacement current density for given magnetic flux density.

Answer to Problem 9.12P

   $J_D=-\frac{B_0{k}_0}{{\mu }_0}\cos \left(\omega t\right)\sin \left(k_{0}z\right)a_x$

Explanation of Solution

Given:

   $B=B_0\cos \left(\omega t\right)\cos \left(k_{0}z\right)a_{y}Wb/m^2$

Medium for magnetic flux density is free space.

Concept used:

Formula for displacement current density is

   $\begin{array}{l}\nabla \times H=J+J_D\left(J_D=\text{displacement}\text{current}\text{density}\right)\\ B={\mu }_{0}H\end{array}$

Calculation:

Plugging the value B in the formula shown below for magnetic field intensity

   $B={\mu }_{0}H$

So,

   $H=\frac{B_0}{{\mu }_0}\cos \left(\omega t\right)\cos \left(k_{0}z\right)a_y$

Hence,

   $\nabla \times H=J+J_D$ ( J is zero for a non-conducting medium)

Therefore,

   $\nabla \times H=J_D$

On calculating curl of H for displace current density

   $\begin{array}{l}{J}_D=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ a_x& a_y& a_z\end{array}\right)\\ J_D=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 0& \frac{B_0}{{\mu }_0}\cos \left(\omega t\right)\cos \left(k_{0}z\right)& 0\end{array}\right)\\ J_D=-\frac{B_0{k}_0}{{\mu }_0}\cos \left(\omega t\right)\sin \left(k_{0}z\right)a_x\end{array}$

To determine

(b)

The electric field intensity.

Answer to Problem 9.12P

   $E=-\frac{B_0{k}_0}{\epsilon {\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)a_x$ volt/m

Explanation of Solution

Given:

   $B=B_0\cos \left(\omega t\right)\cos \left(k_{0}z\right)a_{y}Wb/m^2$

Medium for magnetic flux density is free space.

Concept used:

   $J_D=\frac{\partial D}{\partial t}$

Calculation:

Plugging the value $J_D$ in the formula shown below for magnetic field intensity and integrating with respect to t

So,

   $\begin{array}{l}{\displaystyle \int J_D}dt=-{\displaystyle \int \frac{B_0{k}_0}{{\mu }_0}\cos \left(\omega t\right)\sin \left(k_{0}z\right)dt}\\ D=-\frac{B_0{k}_0}{{\mu }_0}\sin \left(k_{0}z\right){\displaystyle \int \cos \left(\omega t\right)}dt\\ D=-\frac{B_0{k}_0}{{\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)\text{Volt/m}\end{array}$

   $D=\epsilon E$

Hence,

   $E=-\frac{B_0{k}_0}{\epsilon {\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)a_x$ volt/m

To determine

(c)

The value of $k_0$.

Answer to Problem 9.12P

   $k_0=\sqrt{\epsilon {\mu }_0\omega }$

Explanation of Solution

Given:

   $B=B_0\cos \left(\omega t\right)\cos \left(k_{0}z\right)a_{y}Wb/m^2$

Medium for magnetic flux density is free space.

Concept used:

   $\nabla \times E=-\frac{\partial B}{\partial t}$

Calculation:

Plugging the value E in the formula shown above for magnetic flux density

So,

   $\begin{array}{l}-\frac{\partial B}{\partial t}=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ a_x& a_y& a_z\end{array}\right)\\ -\frac{\partial B}{\partial t}=\left(\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ -\frac{B_0{k}_0}{\epsilon {\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)& 0& 0\end{array}\right)\\ \frac{\partial B}{\partial t}=-\frac{B_0{k}_0{}^2}{{\mu }_0}\cos \left(\omega t\right)\sin \left(k_{0}z\right)a_y\end{array}$

On integrating both sides with respect to t

   $\begin{array}{l}{\displaystyle \int \frac{\partial B}{\partial t}}=-{\displaystyle \int \frac{B_0{k}_0{}^2}{{\mu }_0}\cos \left(\omega t\right)\sin \left(k_{0}z\right)}dt\\ B=\frac{B_0{k}_0{}^2}{\epsilon {\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)+c\\ \because c=0\left(\text{since}\text{no}\text{field}\text{are}\text{preasumed}\text{to}\text{exist}\right)\\ \therefore B=\frac{B_0{k}_0{}^2}{\epsilon {\mu }_0\omega }\sin \left(k_{0}z\right)\sin \left(\omega t\right)\mathrm{..................}\left(1\right)\end{array}$

Comparing magnitudes of (1) to the magnitudes of given B

   $B_0$ = $\begin{array}{l}\frac{B_0{k}_0{}^2}{\epsilon {\mu }_0\omega }\\ k_0=\sqrt{\epsilon {\mu }_0\omega }\end{array}$