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.