Question

If the magnetic field of a plane electromagnetic wave is given by (The speed of light =$3\times {10}^8\mathrm{m}{\mathrm{s}}^{-1}$)

$\mathrm{B}=100\times {10}^{-6}\sin \left(2\mathrm{\pi }\times 2\times {10}^{15}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)\right)$then the maximum electric field associated with it is

• A. $4\times {10}^4{\mathrm{NC}}^{-1}$
• B. $6\times {10}^4{\mathrm{NC}}^{-1}$
• C. $3\times {10}^4{\mathrm{NC}}^{-1}$
• D. $4.5\times {10}^4{\mathrm{NC}}^{-1}$

Answer 1

The correct option is C

$3\times {10}^4{\mathrm{NC}}^{-1}$

The explanation for Correct answer:

Option (C) is correct

Step 1 Given data

The magnetic field in a plane electromagnetic wave is given by $\mathrm{B}=100\times {10}^{-6}\sin \left(2\mathrm{\pi }\times 2\times {10}^{15}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)\right)$

Step 2 Formula used

The general magnetic field in a plane electromagnetic wave equation is given by $\mathrm{B}={\mathrm{B}}_0\cos \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{\mathrm{j}}$

The ${\mathrm{E}}_0$ is the electric field vector, and ${\mathrm{B}}_0$ is the magnetic field vector of the Electromagnetic wave is the velocity of light($\mathrm{c}$).

$\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{c}$

Step 3 Electric field

The given magnetic field in a plane electromagnetic wave equation is given by $\mathrm{B}=100\times {10}^{-6}\sin \left(2\mathrm{\pi }\times 2\times {10}^{15}\left(\mathrm{t}-\frac{\mathrm{x}}{\mathrm{c}}\right)\right)$

By comparing the general magnetic field in a plane electromagnetic wave equation is given by $\mathrm{B}={\mathrm{B}}_0\cos \left(\mathrm{kx}-\mathrm{\omega t}\right)\hat{\mathrm{j}}$ with the given equation,

$$\begin{gathered} {\mathrm{B}}_0=100\times {10}^{-6} \\ {\mathrm{B}}_0={10}^{-4} \end{gathered}$$

The velocity of light is, $\mathrm{c}=3\times {10}^8$

The magnetic field vector of the Electromagnetic wave is$\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{c}$

By rearranging this equation, the expression for electric field is,

$$\begin{gathered} {\mathrm{E}}_0=\mathrm{c}\times {\mathrm{B}}_0 \\ {\mathrm{E}}_0=3\times {10}^8\times {10}^{-4} \\ {\mathrm{E}}_0=3\times {10}^4{\mathrm{NC}}^{-1} \end{gathered}$$

Therefore, the maximum electric field associated with the magnetic field of a plane electromagnetic wave is ${\mathrm{E}}_0=3\times {10}^4{\mathrm{NC}}^{-1}$.