Electromagnetic Waves Ques 34
34. The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to :
[2012 M]
(a) the speed of light in vacuum
(b) reciprocal of speed of light in vacuum
(c) the ratio of magnetic permeability to the electric susceptibility of vacuum
(d) unity
Show Answer
Answer:
Correct Answer: 34.(b)
Solution:
- (b) The average energy stored in the electric field
field, $U_E=\frac{1}{2} \varepsilon_0 E^{2}$
The average energy stored in the magnetic field
$=U_B=\frac{1}{2} \frac{B^{2}}{\mu_0}$,
According to conservation of energy $U_E=U_B + W$
$\varepsilon_0 \mu_0=\frac{1}{c^{2}}$
$\frac{B}{E}=\sqrt{\varepsilon_0 \mu_0}=\frac{1}{c}$
The average energy density of an electric field
$\mu_E=\frac{1}{4} \mu_0 E_0^{2}$
$\Rightarrow \mu_E=\frac{1}{4} \in_0(c^{2} B_0^{2}) \quad[\because \frac{E_0}{B_0}=c]$
$\Rightarrow \mu_E=\frac{1}{4} E_0 B_0^{2} \times \frac{1}{\mu_0 \in_0} \quad[\because c=\frac{1}{\sqrt{\mu_0 \in_0}}]$
$\Rightarrow \mu_E=\frac{1}{4} \frac{B_0^{2}}{\mu_0}=\mu_B$
Thus, the average energy density of the electric field equals the average energy density of the magnetic field.
BETA
