Verified Solution • 2025–26 Session moderate

QJEEM23Ph51a6: JEE Main 2023 Physics - Class 12 JEEM23Ph51a6 Solution: The amplitude of magnetic field in an electromagnetic wave…

The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is 6.0×10−7 T6.0 \times 10^{-7} \mathrm{~T} . The maximum value of electric field in the electromagnetic wave is

Multiple Choice Options

  • A 6.0×10−7 Vm−16.0 \times 10^{-7} ~\mathrm{Vm}^{-1}
  • B 5×1014 Vm−15 \times 10^{14} ~\mathrm{Vm}^{-1}
  • C 180 Vm−1180 ~\mathrm{Vm}^{-1}
  • D 2×1015 Vm−12 \times 10^{15} ~\mathrm{Vm}^{-1}

Step-by-Step Verified Solution

In an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation

E=cBE = cB

where c is the speed of light in a vacuum, which is approximately 3×108m/s3 × 10^8 m/s .

Given that the amplitude of the magnetic field B is 6.0×10−7T6.0 × 10^{-7} T , we can substitute these values into the equation to find the maximum value of the electric field:

E=(3×108m/s)×(6.0×10−7T)=180V/mE = (3 × 10^8 m/s) \times (6.0 × 10^{-7} T) = 180 V/m

Therefore, 180 V/m is the correct answer.

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