Question 13 Marks
(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.
(b) Prove that the average energy density of the oscillating electric field is equal to that of the oscillating magnetic field.
Answer
View full question & answer→(a) (i) Radar : Microwaves are used in radar. Frequency range:
$3\times10^{11}$ Hz to $1\times10^{9}$ Hz.
(ii) Eye Surgery : Infrared waves are used Frequency range:
$4\times10^{14}Hz$ to $3\times10^{11}$ Hz.
(b) Energy density in oscilliating electro field
$UE=\frac{1}{2}\in_{0}E^{2}$
Energy density in oscillating magnetic field is :
$UB=\frac{1}{(2\mu_{0})}B^{2}$
As we know $E=CB$
$UE=\frac{1}{2}\in_{0}(CB)^{2}$
$=\frac{1}{2}\in_{0}\frac{1}{\mu_{0}\in_{0}}\times B^{2}=\frac{1}{2\mu_{0}}B^{2}$
$3\times10^{11}$ Hz to $1\times10^{9}$ Hz.
(ii) Eye Surgery : Infrared waves are used Frequency range:
$4\times10^{14}Hz$ to $3\times10^{11}$ Hz.
(b) Energy density in oscilliating electro field
$UE=\frac{1}{2}\in_{0}E^{2}$
Energy density in oscillating magnetic field is :
$UB=\frac{1}{(2\mu_{0})}B^{2}$
As we know $E=CB$
$UE=\frac{1}{2}\in_{0}(CB)^{2}$
$=\frac{1}{2}\in_{0}\frac{1}{\mu_{0}\in_{0}}\times B^{2}=\frac{1}{2\mu_{0}}B^{2}$