# (a) Identify the part of the electromagnetic spectrum used in (i) radar and (ii) eye surgery. Write their frequency range.(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.

(a) (i)Microwaves are used in radar. The frequency range is

$10^{10}\; \; to \; \; 10^{12} Hz$.

(ii) Ultraviolet rays (UV) are used in eye surgery. The frequency range is $10^{15}\; Hz\; \; to \; 10^{17}\; Hz$

(b) We know that the energy density due to the electric field is given by :

$U_E =\frac{1}{2}\; \varepsilon _{o}\; \; E^{2} -----(a)$

Energy density due to the magnetic field is given by :

$U_{B}= \frac{1}{2}\frac{{B^{2}}}{\mu _{o}} ------ (ii)$

where

$\mu _{o}$ = absolute permeability

Hence, we have

$\frac{E}{B}=c$

where c is the speed permeability.

so, $E=cB$

On putting the value of $E_{o}$ in equation (i), we have

$U.E = \frac{1}{2}\; \varepsilon _{o}\; (CB)^{2}$

Here, $C^{2}=\frac{1}{\mu _{o}\varepsilon _{o}}$

then,

$U_E = \frac{1}{2}\varepsilon _{o}\times B^{2}\times \frac{1}{\mu _{o}\varepsilon _{o}}$

$U_E = \frac{B^{2}}{2\mu _{o}} --------(iii)$

Now, on comparing equation (ii) and (iii), we get that average energy density of the oscillating electric field ($U_E$) is equal to that of the oscillating magnetic field.$(U_{B})$.

Hence, $U_{B} = U_{E}= \frac{B^{2}}{2\mu _{o}}$

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