The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is $B_{0}$ = 510 nT. The amplitude of the electric field part of the wave is
120 N $C^{- 1}$
163 N $C^{- 1}$
510 N $C^{- 1}$
153 N $C^{- 1}$

Question

The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is -B-0- - 510 nT- The amplitude of the electric field part of the wave is120 N -C-1-163 N -C-1-510 N -C-1-153 N -C-1-

Toppr Admin · Accepted Answer

Given-B0-510nTc-3-xD7-108m-sThe magnitude of electric field is given byE0-B0cE0-510-xD7-10-x2212-9-xD7-3-xD7-108E0-153NC-x2212-1The correct option is D-

The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. The amplitude of the electric field part of the wave is120 N C−1163 N C−1510 N C−1153 N C−1

Given,B0=510nTc=3×108m/sThe magnitude of electric field is given byE0=B0cE0=510×10−9×3×108E0=153NC−1The correct option is D.

The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is $B_{0}$ = 510 nT. The amplitude of the electric field part of the wave is
120 N $C^{- 1}$
163 N $C^{- 1}$
510 N $C^{- 1}$
153 N $C^{- 1}$

Given,
$B_{0} = 510 n T$
$c = 3 \times 10^{8} m / s$
The magnitude of electric field is given by
$E_{0} = B_{0} c$
$E_{0} = 510 \times 10^{- 9} \times 3 \times 10^{8}$
$E_{0} = 153 N C^{- 1}$
The correct option is D.