Question

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 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 turns/cm = 1500 turns/m
Number of turns per unit length, n = 1500 turns
The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2
Current carried by the solenoid changes from 2 A to 4 A.
∴Change in current in the solenoid, di = 4 − 2 = 2 A
Change in time, dt = 0.1 s
Induced emf in the solenoid is given by Faraday's law as:
$e$=$\frac{d\phi}{dt}$ ...(i)
Where,
$\phi$ = Induced flux through the small loop
= BA... (ii)
B = Magnetic field
=$\mu0ni$ ... (iii)
$\mu$0 = Permeability of free space
= 4nx10-7 H/m
Hence, equation (i) reduces to:

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

= A$\mu_0n$ x $\bigg(\frac{di}{dt}\bigg)$
=2×10-4×4π×10-7x1500 x $\frac{2}{0.1}$
= 7.54×10-6 V
Hence, the induced voltage in the loop is 7.54x10-6 V.