Electromagnetic Waves - Result Question 9

9. The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to :

(a) the speed of light in vacuum [2012M]

(b) reciprocal of speed of light in vacuum

(c) the ratio of magnetic permeability to the electric susceptibility of vacuum

(d) unity

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Answer:

Correct Answer: 9. (b)

Solution:

  1. (b) The average energy stored in the electric

field, UE=12ε0E2

The average energy stored in the magnetic field

=UB=12B2μ0,

According to conservation of energy UE=UB ε0μ0=B2E2

BE=ε0μ0=1c

The average energy density of electric field

μE=140E02

μE=140(c2B02)[E0B0=c]

μE=14E0B02×1μ00[c=1μ00]

μE=14B02μ0=μB

Thus, the average energy density of electric field equal to the average density of magnetic field.



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