A long solenoid with 15 turns per $\mathrm{cm}$ has a small loop of area $2.0 \mathrm{~cm}^2$ placed inside…
A long solenoid with 15 turns per $\mathrm{cm}$ has a small loop of area $2.0 \mathrm{~cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 Ato $4.0 \mathrm{~A}$ in $0.1 \mathrm{~s}$, the induced emf in the loop while the current is changing is nearly [Take $\pi=3.14$ ]
$9 \times 10^{-6} \mathrm{~V}$
$4.48 \times 10^{-6} \mathrm{~V}$
$5.2 \times 10^{-6} \mathrm{~V}$
$7.54 \times 10^{-6} \mathrm{~V}$
Solution
The magnetic field produced inside the solenoid is $B=\mu_0 n I$
If $A$ is the area of the loop placed inside the solenoid, then the magnetic flux linked with the loop is given by,
$\phi=B A=\mu_0 n I A$
If $e$ is the induced e.m.f. produced due to change in current through the solenoid, then
$e=-\frac{d \phi}{d t}=-\frac{d}{d t}\left[\mu_0 n I A\right]=-\mu_0 \times n \times A \times \frac{d I}{d t}$
Given: Number of turns per unit length of the solenoid,
$n=15$ turns $\mathrm{cm}^{-1}=1500$ turns $\mathrm{m}^{-1}$
Given: $A=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2$ and $\frac{d I}{d t}=\frac{4-2}{0.1}=20 \mathrm{As}^{-1}$
$\therefore e=-4 \pi \times 10^{-7} \times 1500 \times 2 \times 10^{-4} \times 20=-7.54 \times 10^{-6} \mathrm{~V}$.