Question

If the magnetic field in a plane electromagnetic wave is given by $\mathrm{B}=3\times {10}^{-8}\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{j}}$ $\mathrm{T}$ then what will be the expression for the electric field?

• A. $\mathrm{E}=3\times {10}^{-8}\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{i}}\mathrm{V}{\mathrm{m}}^{-1}$
• B. $\mathrm{E}=3\times {10}^{-8}\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{j}}\mathrm{V}{\mathrm{m}}^{-1}$
• C. $\mathrm{E}=60\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{k}}\mathrm{V}{\mathrm{m}}^{-1}$
• D. $\mathrm{E}=9\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{k}}\mathrm{V}{\mathrm{m}}^{-1}$

Answer 1

The correct option is D

$\mathrm{E}=9\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{k}}\mathrm{V}{\mathrm{m}}^{-1}$

The explanation for the correct answer:

Option (D) is correct

Step 1 Given data

The magnetic field in a plane electromagnetic wave is given by $\mathrm{B}=3\times {10}^{-8}\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{j}}$ $\mathrm{T}$

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

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

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

Step 3 Electric field

The given magnetic field in a plane electromagnetic wave equation is given by $\mathrm{B}=3\times {10}^{-8}\sin (1.6\times {10}^3\mathrm{x}+48\times {10}^{10}\mathrm{t})\hat{\mathrm{j}}$ $\mathrm{T}$

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,

$$\begin{gathered} {\mathrm{B}}_0=3\times {10}^{-8} \\ \mathrm{k}=1.6\times {10}^3 \\ \mathrm{\omega }=48\times {10}^{10} \end{gathered}$$

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,

$$\begin{gathered} {\mathrm{E}}_0=\mathrm{c}\times {\mathrm{B}}_0 \\ {\mathrm{E}}_0=3\times {10}^8\times 3\times {10}^{-8} \\ {\mathrm{E}}_0=9\mathrm{V}{\mathrm{m}}^{-1} \end{gathered}$$

The expression electromagnetic wave equation is given by $\mathrm{E}={\mathrm{E}}_0\cos \left(\mathrm{kx}-\mathrm{\omega t}\right)$

That is,

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_0\cos \left(\mathrm{kx}-\mathrm{\omega t}\right) \\ \mathrm{E}=9\cos \left(1.6\times {10}^3\mathrm{x}-48\times {10}^{10}\mathrm{t}\right) \end{gathered}$$

Therefore, the expression electromagnetic wave equation is given by $\mathrm{E}=9\cos \left(1.6\times {10}^3\mathrm{x}-48\times {10}^{10}\mathrm{t}\right)\hat{\mathrm{k}}$

Hence, option (D) is the correct answer.