Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

The magnetic field in a travelling electromagnetic wave has a peak value of 30 nT. The peak value of electric field strength is: (speed of light in free space is $3 \times 10^8 m/s$).

Options:

9.0 V/m

3.0 V/m

6.0 V/m

18.0 V/m

Correct Answer:

9.0 V/m

Explanation:

The correct answer is Option (1) → 9.0 V/m

$E_{max}=cB_{max}$

Where,

$E_{max}$ = Peak value of electric field

$B_{max}$ = Peak value of magnetic field

$c = 3×10^8m/s$ (Speed of light)

$E_{max}=(3×10^8)(30×10^{-9})$

$=9V/m$