Question

A long solenoid with $20$ turns per $cm$ has a small loop of area $2.0c{m}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0A$ to $4.0A$ in $0.1sec$, calculate the induced $emf$ in the loop while the current is changing.

Answer 1

Number of turns on the solenoid $=$ 15 turns/cm $=$ 1500 turns/m

Number of turns per unit length, n $=$ 1500 turns

The solenoid has a small loop of area, A $=2.0c{m}^2$

Current carried by the solenoid changes from 2 A to 4 A.

Change in current in the solenoid, $di=4-2=2A$

Change in time, $dt=0.1s$

Induced emf in the solenoid is given by Faraday’s law as:

$e=\frac{d\varphi }{dt}$ ...(i)

Where,

$\varphi =$ Induced flux through the small loop

$=BA$ ... (ii)

$B=$ Magnetic field

$={\mu }_{0}ni$ ...(iii)

$\mu 0=$ Permeability of free space

$=4\pi \times {10}^{-7}H/m$

Hence, equation (i) reduces to:

$e=\frac{d}{dt}\left(BA\right)$

$=A{\mu }_{0}n\times \left(\frac{di}{dt}\right)$

$=2\times {10}^{-4}\times 4\pi \times {10}^{-7}\times 1500\times \frac{2}{0.1}$

$=7.54\times {10}^{-6}V$

Hence, the induced voltage in the loop is $7.54\times {10}^{-6}V$.