A small coil of 10 turns is placed inside a solenoid of length 20 cm and 240 turns carry a current of $\frac{10}{\mathrm{\pi }}$ A The area of small coil is 2.5 cm² and resistance 4.8 $\mathrm{\Omega }$ then the current reduces to zero in 25 ms, the value of average induced current is :

Values given for coil :

Number of turns of small coil = 10

Area of coil= 2.5 cm² = 2.5 x 10^-4m²

$∆\mathrm{t}=25 \mathrm{x} {10}^{-3}\mathrm{s}$

Values given for solenoid

I = 20 x 10^-2 m, N = 240

$\mathrm{I}=\frac{10}{\mathrm{\pi }} \mathrm{amp}, \mathrm{n}=\frac{240}{20 \mathrm{x} {10}^{-2}}=12 \mathrm{x} {10}^2$

According to Faraday's law, induced emf $\mathrm{E}=-\frac{\mathrm{d\varphi }}{\mathrm{dt}}$

Induced emf in coil :

$$\begin{gathered} \mathrm{I}=\frac{\left|-{\displaystyle \frac{\mathrm{d\varphi }}{\mathrm{dt}}}\right|}{\mathrm{R}}=\frac{\left(\mathrm{NBA}\right)/ \mathrm{dt}}{\mathrm{R}}=\frac{{\mathrm{\mu }}_0\mathrm{nI}\left(\mathrm{NA}\right)}{∆\mathrm{tR}} \\ =\frac{10 \mathrm{x} 4\mathrm{\pi } \mathrm{x} {10}^{-7} \mathrm{x} 12 \mathrm{x} {10}^2 \mathrm{x} 10/\mathrm{\pi } \mathrm{x} 2.5 \mathrm{x} {10}^{-4}}{25 \mathrm{x} {10}^{-3} \mathrm{x} 4.8} \\ =\frac{4 \mathrm{x} 12 \mathrm{x} {10}^{-7} \mathrm{x} {10}^2 \mathrm{x} 10 \mathrm{x} {10}^{-4}}{{10}^{-3} \mathrm{x} 4.8}={10}^{-4} \\ \mathrm{I}=0.1 \mathrm{mA} \end{gathered}$$