**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?

**Solution:**

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:

$e=\frac{d \phi}{d t}$ ....(i)

Where,

$\phi=$ Induced flux through the small loop

= *BA *... (*ii*)

*B *= Magnetic field

$=\mu_{0} n i$ ...(iii)

μ0 = Permeability of free space

*= *4π×10−7 H/m

Hence, equation (*i*) reduces to:

$e=\frac{d}{d t}(B A)$

$=A \mu_{0} n \times\left(\frac{d i}{d t}\right)$

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

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

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

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