CBSE Class 12 · 2019 · EM WavesShort Answer

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(a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range. (b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.

Solution

(a) (i) Radar uses microwaves, of frequency about

1

GHz to

10

GHz. (ii) Eye surgery uses ultraviolet rays, of frequency about

1 0^{15}

1 0^{17}

Hz.
(b) For a plane EM wave

E = E_{0} sin (k x - ω t)

B = B_{0} sin (k x - ω t)

.
Average electric energy density:

u_{E} = ⟨ \frac{1}{2} ε_{0} E^{2} ⟩ = \frac{1}{2} ε_{0} E_{0}^{2} ⟨ sin^{2} ⟩ = \frac{1}{4} ε_{0} E_{0}^{2}

Average magnetic energy density:

u_{B} = ⟨ \frac{B ^{2}}{2 μ _{0}} ⟩ = \frac{B _{0}^{2}}{4 μ _{0}}

Using

B_{0} = E_{0} / c

and

c^{2} = 1/ (μ_{0} ε_{0})

u_{B} = \frac{E _{0}^{2}}{4 μ _{0} c ^{2}} = \frac{μ _{0} ε _{0} E _{0}^{2}}{4 μ _{0}} = \frac{1}{4} ε_{0} E_{0}^{2} = u_{E}

u_{E} = u_{B}

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CBSE Class 12 · 2019 · EM WavesShort Answer

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