JEE Main 2022 — Electromagnetic Waves Question with Solution

Given the electric field expression:

$E_z=300\sin (5\pi \times 10^{3}x-3\pi \times 10^{11}t){\text{Vm}}^{-1}$

The amplitude of the electric field ($E_0$) is $300\text{V/m}$.

The velocity ($v$) of the wave in the medium is given by the ratio of the coefficients of time and displacement in the wave equation, which can be calculated as:

$v=\frac{\text{Coefficient of }t}{\text{Coefficient of }x}=\frac{3\pi \times 10^{11}}{5\pi \times 10^3}=\frac{3}{5}\times 10^8\text{m/s}$

The relationship between the electric field amplitude and the magnetic field amplitude in an electromagnetic wave is given by:

$B_0=\frac{E_0}{v}$

Substituting the values for $E_0$ and $v$ into this formula:

$B_0=\frac{300}{\frac{3}{5}\times 10^8}$

$B_0=\frac{300\times 5}{3\times 10^8}$

$B_0=\frac{1500}{3\times 10^8}$

$B_0=5\times 10^{-6}\text{T}$

Therefore, the amplitude of the magnetic field ($B_0$) is $5\times 10^{-6}$ T.