Verified Solution • 2025–26 Session moderate

QJEEM20Ph0132: JEE Main 2020 Physics - Class 12 JEEM20Ph0132 Solution: The magnetic field of a plane electromagnetic wave is T where c …

The magnetic field of a plane electromagnetic wave is
B→=3×10−8sin⁡[200π(y+ct)]i^\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i T
where c = 3 ×\times 108 ms–1 is the speed of light. The corresponding electric field is :

Multiple Choice Options

  • A E→=−10−6sin⁡[200π(y+ct)]k^\overrightarrow E = - {10^{ - 6}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k V/m
  • B E→=−9sin⁡[200π(y+ct)]k^\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k V/m
  • C E→=9sin⁡[200π(y+ct)]k^\overrightarrow E = 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k V/m
  • D E→=3×10−8sin⁡[200π(y+ct)]k^\overrightarrow E = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k

Step-by-Step Verified Solution

Given, B→=3×10−8sin⁡[200π(y+ct)]i^\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i T

∴\therefore E0 = CB0

= 3 × 108 × 3 × 10–8 = 9 V/m

We know, (E→×B→)∣∣C→\left( {\overrightarrow E \times \overrightarrow B } \right)||\overrightarrow C

And here B^=i^&C^=−j^\widehat B = \widehat i\& \widehat C = - \widehat j

∴\therefore E^=−k^\widehat E = - \widehat k

So, E→=−9sin⁡[200π(y+ct)]k^\overrightarrow E = - 9\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat k V/m

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