QJEEM22Ph8282: JEE Main 2022 Physics - Class 12 JEEM22Ph8282 Solution: The oscillating magnetic field in a plane electromagnetic wave…

In an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :

$$E=Bv$$

Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave.

In the given problem, the peak value of the magnetic field is given as :

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

The wave speed (v) can be calculated from the given equation for the magnetic field, using the relationship between frequency (f), wave number (k), and speed :

$$v=\frac{\omega }{k}=\frac{1000\pi \times 4\times {10}^8}{1000\pi \times 5}=8\times {10}^{7}m/s$$

Substituting these values into the equation for the electric field gives the peak value of the electric field :

$$E_0=B_{0}v=5\times {10}^{-6}T\times 8\times {10}^{7}m/s=4\times {10}^{2}V/m$$