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Home » The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is . What is the amplitude of the electric field part of the wave?
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is \mathbf{B}_{0}=510 \mathrm{nT}. What is the amplitude of the electric field part of the wave?
CBSE | Class 12 | Electromagnetic Waves | NCERT | Physics
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is \mathbf{B}_{0}=510 \mathrm{nT}. What is the amplitude of the electric field part of the wave?

Amplitude of magnetic field of an electromagnetic wave in a vacuumis given to us as,

B_{0}=510 n T=510 \times 10^{-9} T

Speed of light in vacuum is known as \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}

The relation for the amplitude of electric field of an electromagnetic wave is,

E=c B_{0}=3 \times 10^{8} \times 510 \times 10^{-9}=153 \mathrm{~N} / C

As a result, 153 \mathrm{~N} / \mathrm{C} is the electric field part of the wave.

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