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NCERT Class XII Chapter
Electromagnetic Waves
Questions With Solutions

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Question : 10
Total: 15
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m–1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the
→
E
 field equals the average energy density of the
→
B
 field. [c = 3 × 108 m s–1.]
Solution:  
(a) Wavelength of electromagnetic wave is
λ =
c
v
 =
3×108
2×1010
 = 1.5 × 10−2 m
(b) Amplitude of magnetic field,
E0
B0
 = c
B0 =
E0
c
 =
48
3×108
 = 16 × 10−8 T
(c) Energy density as electric field, uE =
1
2
ε0E2
Here,
E
B
 = c
So, uE =
1
2
ε0c2B2
where speed of electromagnetic wave, c =
1
√µ0ε0
so uE =
1
2
ε0
µ0ε0
B2 =
B2
2µ0
 = µB
[Energy density as magnetic field]
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