Verified Solution • 2025–26 Session moderate

QJEEM22Phb853: JEE Main 2022 Physics - Class 12 JEEM22Phb853 Solution: The magnetic field of a plane electromagnetic wave is given by…

The magnetic field of a plane electromagnetic wave is given by :

B→=2×10−8sin⁡(0.5×103x+1.5×1011t) j^ T\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T} .

The amplitude of the electric field would be :

Multiple Choice Options

  • A 6 Vm−16\, \mathrm{Vm}^{-1} along xx -axis
  • B 3 Vm−13\, \mathrm{Vm}^{-1} along zz -axis
  • C 6 Vm−16\, \mathrm{Vm}^{-1} along zz -axis
  • D 2×10−8 Vm−12 \times 10^{-8} \,\mathrm{Vm}^{-1} along zz -axis

Step-by-Step Verified Solution

Speed of light c=ωk=1.5×10110.5×103=3×108c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8} m/sec

So, E0=B0c{E_0} = {B_0}c

=2×10−8×3×108= 2 \times {10^{ - 8}} \times 3 \times {10^8}

=6= 6 V/m

Direction will be along z-axis.

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