Q71 / 90 · JEE Main 2023 (11 Apr Shift 2) · Physics

A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular space and time E→=6.6j ^ V m-1. What is B→ at this point?

  • Correct answer
    A
    2.2×10-8k^T
  • B
    -2.2×10-8i^T
  • C
    -2.2×10-8k^T
  • D
    2.2×10-8i ^T

Solution

It is given that 

 E→=6.6j^( V m-1).

The relation between the electric, magnetic field and speed of light is E→×B→=c→.
Figure
Clearly,   B→=B0(k^)

Using the relation,
     cB0=E
⇒B0=Ec=6.63×108=2.2×10-8 T

Hence, the correct answer is option: 'A'.

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