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:
- (b) The average energy stored in the electric
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 $U_E=U_B$ $\varepsilon_0 \mu_0=\frac{B^{2}}{E^{2}}$
$\frac{B}{E}=\sqrt{\varepsilon_0 \mu_0}=\frac{1}{c}$
The average energy density of electric field
$\mu_E=\frac{1}{4} \in_0 E_0^{2}$
$\Rightarrow \mu_E=\frac{1}{4} \in_0(c^{2} B_0^{2}) \quad[\because \frac{E_0}{B_0}=c]$
$\Rightarrow \mu_E=\frac{1}{4} E_0 B_0^{2} \times \frac{1}{\mu_0 \in_0} \quad[\because c=\frac{1}{\sqrt{\mu_0 \in_0}}]$
$\Rightarrow \mu_E=\frac{1}{4} \frac{B_0^{2}}{\mu_0}=\mu_B$
Thus, the average energy density of electric field equal to the average density of magnetic field.
