Question

A very long solenoid has $15$ turns per $\text{cm}$ and a small loop of area $2{\text{cm}}^2$ is placed in this solenoid, perpendicular to its axis. The induced emf in the loop while the current carried by the solenoid is changing steadily from $2.0\text{A}$ to $4.0\text{A}$ in $0.1\text{s}$ is $7.54\times {10}^{-x}\text{V}$. Then, find the value of $x$

Answer 1

Number of turns on the solenoid $=15\text{turns/cm}=1500\text{turns/m}$
Number of turns per unit length, $n=1500$ turns.
The solenoid has a small loop of area, $A=2{\text{cm}}^2=2\times {10}^{-4}{\text{m}}^2$
Current carried by the solenoid changes from $2.0\text{A}$ to $4.0\text{A}$.
Therefore, change in current in the solenoid is$=4-2=2\text{A}$
Change in time, $dt=0.1\text{s}$
According to Faraday's law induced emf in the solenoid is given by
$e=\frac{d\varphi }{dt}.....\left(1\right)$
where $\varphi$ is flux induced in small loop$=BA$
$B=\text{Magnetic field}={\mu }_{0}ni$
Hence,
$e=\frac{d\varphi }{dt}=A{\mu }_{0}n\frac{di}{dt}=2\times {10}^{-4}\times 4\pi \times {10}^{-7}\times 1500\times \frac{2}{0.1}$
$e=7.54\times {10}^{-6}\text{V}$
Hence, $x=6$