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:

  1. 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, $\quad U _E=U _B$



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