A long solenoid with 2000 turns per meter has a small loop of radius \(3 \mathrm{~cm}\) placed inside the…
A long solenoid with 2000 turns per meter has a small loop of radius \(3 \mathrm{~cm}\) placed inside the solenoid normal to its axis. If the current through the solenoid increases steadily from \(1.5 \mathrm{~A}\) to \(5.5 \mathrm{~A}\) in \(\frac{\pi^2}{100} \mathrm{~s}\), the induced emf in the loop is
\(0.144 \mathrm{mV}\)
\(0.288 \mathrm{mV}\)
\(0.072 \mathrm{mV}\)
\(0.316 \mathrm{mV}\)
Solution
Key idea Magnetic field inside a solenoid of infinite length is given by expression
\(B=\mu_0 n i\)
where, \(n=\) number of turns per unit length.
Given, number of turns in solenoid, \(n=2000\), current through solenoid, \(i_i=1.5 \mathrm{~A}\) and \(i_f=5.5 \mathrm{~A}\)
So, \(\Delta B=\mu_0 n\left(i_f-i_i\right)\)
Putting the given values,
\(\begin{aligned}
& =4 \pi \times 10^{-7} \times 2000(5.5-1.5) \\
\Rightarrow \quad \Delta B & =4 \pi \times 10^{-7} \times 2000 \times 4
\end{aligned}\)
Now, emf induced in a loop of radius \(3 \mathrm{~cm}\),
\(\begin{gathered}
e=-\frac{d \phi}{d t}=\frac{\Delta B A}{\Delta t} \cos \theta \\
\Rightarrow e=\frac{4 \pi \times 10^{-7} \times 2000 \times 4 \times \pi \times\left(3 \times 10^{-2}\right)^2\left(\cos 0^{\circ}\right)}{\frac{\pi^2}{100}} \\
\Rightarrow E=0.288 \mathrm{mV} \\
\quad\left(\because \theta=0^{\circ}\right)
\end{gathered}\)
Hence, the correct option is (b).