Question

The oscillating magnetic field of a plane electromagnetic wave is given as $B=4\times {10}^{–6}\sin [200\pi x–30\times {10}^9\pi t]$ Tesla. The amplitude of electric field is

• A. $90\text{V/m}$
• B. $600\text{V/m}$
• C. $900\text{V/m}$
• D. $1200\text{V/m}$

Answer 1

The correct option is B $600\text{V/m}$

Comparing the given equation from general equation i.e. $B=B_0\sin (kx-\omega t)$, we have
$B_0=4\times {10}^{-6},k=200\pi ,\omega =30\times {10}^9\pi$
$As\frac{E}{B_0}=v\text{where,}v\text{is speed of wave and}v=\frac{\omega }{k}=\frac{30\times {10}^9\pi }{200\pi }=1.5\times {10}^8\text{m/s}$
Thus,
$E=v\times B_0$
$\Rightarrow E=1.5\times {10}^8\times 4\times {10}^{-6}=600\text{V/m}$

Hence, option $\left(D\right)$ is the correct choice.