Choose the correct answer from the alternatives given.
The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is $B_{0} = 510 n T .$ The amplitude of the electric field aprt of the wave is:

120 N C^{- 1}

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134 N C^{- 1}

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5100 N C^{- 1}

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153 N C^{- 1}

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Solution

The correct option is B $153 N C^{- 1}$
Given: The magnetic field at the particular point is $510 n T$

To find: The electric field at that point.

The electric field of the electromagnetic wave at a point is given by:
$\frac{E}{B} = C \Rightarrow E = C B = 3 \times 10^{8} \times \frac{510}{10^{9}} = 153 N / C$

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Choose the correct answer from the alternatives given.The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is B0=510nT. The amplitude of the electric field aprt of the wave is:

The correct option is B 153NC−1Given: The magnetic field at the particular point is 510 nTTo find: The electric field at that point.The electric field of the electromagnetic wave at a point is given by:EB=C⇒E=CB=3×108×510109=153N/C

Choose the correct answer from the alternatives given.
The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is $B_{0} = 510 n T .$ The amplitude of the electric field aprt of the wave is: