Question

A plane electromagnetic wave propagating in the $x-$direction in vacuum has wavelength of $6.0\text{mm}$. The electric field is in the $y-$direction and its maximum magnitude is $33{\text{Vm}}^{-1}$.

$\left(ii\right)$ The equation for the magnetic field as a function of $x\left(\text{m}\right)$ and $t\left(\text{s}\right)$ is

​​​​​​​$[$1 Mark$]$

• A. $\overrightarrow{B}=11\times {10}^{-8}\sin \left[\pi (t-\frac{x}{c})\right]\hat{i}$
• B. $\overrightarrow{B}=11\times {10}^{-8}\sin [\pi \times {10}^{11}(t-\frac{x}{c})]\hat{i}$
• C. $\overrightarrow{B}=11\times {10}^{-8}\sin [\pi \times {10}^{11}(t-\frac{x}{c})]\hat{j}$
• D. $\overrightarrow{B}=11\times {10}^{-8}\sin [\pi \times {10}^{11}(t-\frac{x}{c})]\hat{k}$

Answer 1

The correct option is D $\overrightarrow{B}=11\times {10}^{-8}\sin [\pi \times {10}^{11}(t-\frac{x}{c})]\hat{k}$

Given, $\lambda =6\text{mm}=6\times {10}^{-3}\text{m}$

The direction of propagation of electromagnetic wave is given by $\overrightarrow{V}=\overrightarrow{E}\times \overrightarrow{B}$

Here, direction of $\overrightarrow{V}$ is along x-axis (given)
direction of $\overrightarrow{E}$ is along y-axis (given).

So the direction of $\overrightarrow{B}$ should be along z-axis.

The equation for the magnetic field along z-axis of the electromagnetic wave is given by,

$\overrightarrow{B}=B_0\sin \omega (t-\frac{x}{c})\hat{k}$

$B_0=\frac{E_0}{c}=\frac{33}{3\times {10}^8}=11\times {10}^{-8}\text{T}$

Now, $\omega =2\pi f=\frac{2\pi c}{\lambda }$
$\omega =\frac{2\pi \times 3\times {10}^8}{6\times {10}^{-3}}=\pi \times {10}^{11}$

So, $\overrightarrow{B}=11\times {10}^{-8}\sin [\pi \times {10}^{11}(t-\frac{x}{c})]\hat{k}$