Verified Solution • 2025–26 Session moderate

QJEEM25Ph8cf1: JEE Main 2025 Physics - Class 12 JEEM25Ph8cf1 Solution: A plane electromagnetic wave of frequency 20 MHz travels in…

A plane electromagnetic wave of frequency 20 MHz travels in free space along the +x+x direction. At a particular point in space and time, the electric field vector of the wave is Ey=9.3Vm−1\mathrm{E}_y=9.3 \mathrm{Vm}^{-1} . Then, the magnetic field vector of the wave at that point is

Multiple Choice Options

  • A Bz=1.55×10−8 T\mathrm{B}_z=1.55 \times 10^{-8} \mathrm{~T}
  • B Bz=6.2×10−8 T\mathrm{B}_z=6.2 \times 10^{-8} \mathrm{~T}
  • C Bz=3.1×10−8 T\mathrm{B}_z=3.1 \times 10^{-8} \mathrm{~T}
  • D Bz=9.3×10−8 T\mathrm{B}_z=9.3 \times 10^{-8} \mathrm{~T}

Step-by-Step Verified Solution

For an electromagnetic wave in free space, the magnitudes of the electric and magnetic fields are related by the speed of light, c≈3×108m/s, through the equation

E=cB.

Here's how to determine the magnetic field:

We are given the electric field magnitude: E=9.3V/m.

Use the relationship to find the magnetic field magnitude:

B=Ec=9.33×108T.

Calculating the value gives:

B≈3.1×10−8T.

Since the electric field is along the y direction and the wave propagates along the +x direction, by the right-hand rule, the magnetic field is oriented along the z direction.

Thus, the magnetic field vector is

Bz=3.1×10−8T,

which corresponds to Option C.

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