Question

A long solenoid with 15 turns per cm has a small loop of area $2.0{\text{cm}}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A 𝑡𝑜 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

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,
$A=2.0{\text{cm}}^2=2\times {10}^{-4}{m}^2$
The current carried by the solenoid changes from 2 A to 4 A.
Therefore, change in current in the solenoid,
$di=4-2=2A$
Change in time, $dt=0.1s$
Therefore, rate of change in current, $\frac{di}{dt}=20A/s$
According to Faraday's law, induced emf in the solenoid is given by:
$e=\frac{d\varphi }{dt}....\left(i\right)$ Where,
$\varphi$ = induced flux through the small loop
$\varphi =BA....\left(ii\right)$ B = Magnetic field due to a solenoid
$B={\mu }_{0}ni{\mu }_0$ = Permeability of free space
${\mu }_0=4\pi \times {10}^{-7}\frac{H}{m}$
Hence, equation (i) can be rduced to:
$e=\frac{d}{dt}\left(BA\right)$
Put $B={\mu }_{0}ni$ in the given equation,
$e=A{\mu }_{0}n\times \left(\frac{di}{dt}\right)$
$e=2\times {10}^{-4}\times 4\pi \times {10}^{-7}\times 1500\times \frac{2}{0.1}$
$e=7.54\times {10}^{-6}v$ Hence, the induced voltage in the loop is $7.54\times {10}^{-6}V$
Final answer: $7.5\times {10}^{-6}V$