First slide

Electromagnetic waves and its nature

Question

The peak value of electric field produced by the radiation coming from a $100 w a t t$ bulb at a distance of $3 m$ is ( assume that the efficiency of the bulb is 2.5% and it is a point source)

Moderate

A

$10 V m^{- 1}$
B

$4.08 V m^{- 1}$
C

$6 V m^{- 1}$
D

$8 V m^{- 1}$

Solution

Useful intensity, $I = \frac{p o w e r}{a r e a} = \frac{100 \times \frac{2.5}{100}}{4 π {(3)}^{2}}$
$\Rightarrow I = \frac{2.5}{36 π} W m^{- 2}$
Half of this intensity(I) belongs to electric field and half of that to magnetic field
$\frac{I}{2} = \frac{1}{4} \in_{0} E_{0}^{2} C \Rightarrow E_{0} = \sqrt{\frac{2 I}{\in_{0} C}}$
$\Rightarrow E_{0} = \sqrt{\frac{2 \times (2.5 / 36 π)}{(\frac{1}{4 π \times 9 \times 10^{9}}) \times (3 \times 10^{8})}} = 4.08 V m^{- 1}$

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