Modern Physics 7 Question 27
28. In free space, the energy of electromagnetic wave in electric field is $U _E$ and in magnetic field is $U _B$. Then
(a) $U _E=U _B$
(b) $U _E>U _B$
(c) $U _E<U _B$
(d) $U _E=\frac{U _B}{2}$
(Main 2019, 9 Jan II)
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Solution:
- Energy density of an electomagnetic wave in electric field,
$$ U _E=\frac{1}{2} \varepsilon _0 \cdot E^{2} $$
Energy density of an electromagnetic wave in magnetic field,
where, $E=$ electric field,
$$ U _B=\frac{B^{2}}{2 \mu _0} $$
$B$ =magnetic field,
$\varepsilon _0=$ permittivity of medium and
$\mu _0=$ magnetic permeability of medium.
From the theory of electro-magnetic waves, the relation between $\mu _0$ and $\varepsilon _0$ is
$$ c=\frac{1}{\sqrt{\mu _0 \varepsilon _0}} $$
where, $c=$ velocity of light $=3 \times 10^{8} m / s$
$$ \text { and } \quad \frac{E}{B}=c $$
Dividing Eq. (i) by Eq. (ii), we get
$$ \frac{U _E}{U _B}=\frac{\frac{1}{2} \varepsilon _0 E^{2}}{\frac{1}{2} B^{2} \times \frac{1}{\mu _0}}=\frac{\mu _0 \varepsilon _0 E^{2}}{B^{2}} $$
Using Eqs. (iii), (iv) and (v), we get
$$ \frac{U _E}{U _B}=\frac{c^{2}}{c^{2}}=1 $$
Therefore, $\quad U _E=U _B$
