Question

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

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

Answer 1

The correct option is D $\overrightarrow{E}=9\sin (1.6\times {10}^{3}x+48\times {10}^{10}t)\hat{k}\text{V/m}$

We know that,
$$\begin{array}{cc}& \left|\frac{E_0}{B_0}\right|=c\\ & \text{Given}{B}_0=3\times {10}^{-8}\\ & \Rightarrow E_0=B_0\times c=3\times {10}^{-8}\times 3\times {10}^8\\ & \Rightarrow E_0=9\text{N/C}\\ & \therefore E=E_o\sin (\omega t-kx+\varphi )\hat{k}=9\sin (\omega t-kx+\varphi )\hat{k}\\ & \overrightarrow{E}=9\sin (1.6\times {10}^{3}x+48\times {10}^{10}t)\hat{k}\text{V/m}\end{array}$$