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An air-cored solenoid with length $30 \mathrm{~cm}$, area of cross-section $25 \mathrm{~cm}^2$ and number of turns 500 carries a current of $2.5 \mathrm{~A}$. The current is suddenly switched off for a brief time of $10^{-3} \mathrm{~s}$. How much is the (nearly) average back e.m.f. induced across the ends of the open switch in the circuit?
(Ignore the variation of magnetic field near the ends of solenoid)
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(Ignore the variation of magnetic field near the ends of solenoid)
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2433 Upvotes
Verified Answer
The correct answer is:
6.5 V
Field at the centre of the solenoid with $N$ turns over length $l$ is :
$$
B=\mu_0\left(\frac{N}{l}\right) i
$$
The self-flux associated with the coil with area $A$ is: $\Phi=N(B A)$ We define self-inductance $L$ as, $\Phi=L i$
Therefore, self-inductance of the solenoid is given by,
$$
L=\frac{\Phi}{i}=\frac{N(B A)}{i}=\frac{N\left[\mu_0\left(\frac{N}{l}\right) i A\right]}{i}=\frac{\mu_0 N^2 A}{l}
$$
Back emf,
$$
\begin{aligned}
& e=L\left(\frac{d i}{d t}\right)=\left(\frac{\mu_0 N^2 A}{l}\right) \frac{d i}{d t}=\frac{4 \pi \times 10^{-7} \times 500 \times 500 \times 25 \times 10^{-4}}{0.3} \times \frac{2.5}{10^{-3}} \mathrm{~V} \\
& \therefore e=6.5 \mathrm{~V}
\end{aligned}
$$
$$
B=\mu_0\left(\frac{N}{l}\right) i
$$
The self-flux associated with the coil with area $A$ is: $\Phi=N(B A)$ We define self-inductance $L$ as, $\Phi=L i$
Therefore, self-inductance of the solenoid is given by,
$$
L=\frac{\Phi}{i}=\frac{N(B A)}{i}=\frac{N\left[\mu_0\left(\frac{N}{l}\right) i A\right]}{i}=\frac{\mu_0 N^2 A}{l}
$$
Back emf,
$$
\begin{aligned}
& e=L\left(\frac{d i}{d t}\right)=\left(\frac{\mu_0 N^2 A}{l}\right) \frac{d i}{d t}=\frac{4 \pi \times 10^{-7} \times 500 \times 500 \times 25 \times 10^{-4}}{0.3} \times \frac{2.5}{10^{-3}} \mathrm{~V} \\
& \therefore e=6.5 \mathrm{~V}
\end{aligned}
$$
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