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A long solenoid $\mathrm{S}$ has $\mathrm{n}$ turns per meter, with diameter a. At the centre of this coil, we place a smaller coil of $\mathrm{N}$ turns and diameter $\mathrm{b}$ (where $\mathrm{b} < \mathrm{a}$ ). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of $\mathrm{mt}^2+\mathrm{C}$.
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Verified Answer
According to the question, when verying current is passed through solenoid the verying magnetic field can induce the current in another smaller coil.
So, magnetic field due to a solenoid $S_1$ is
$$
\mathrm{B}=\mu_0 \mathrm{n} l
$$
Magnetic flux in smaller coil
$$
\begin{aligned}
\phi &=\mathrm{NBA} \\
\mathrm{A} &=\pi \mathrm{b}^2
\end{aligned}
$$
By applying Faraday's law of EMI, the induced emf in coil due to solenoids verying magnetic field.
$$
e=\frac{-d \phi}{d t} \quad(\because \phi=N B A)
$$
So, $e=\frac{-d}{d t}(\mathrm{NBA}) \quad\left(\therefore \mathrm{B}=\mu_0 \mathrm{Ni}\right)$
$$
\begin{aligned}
\mathrm{e}=-\mathrm{N} \pi \mathrm{b}^2 \mu_0 \mathrm{n} \frac{\mathrm{dl}}{\mathrm{dt}} &=-\mathrm{Nn} \pi \mu_0 \mathrm{~b}^2 \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{mt}^2+\mathrm{C}\right) \\
&=-\mu \mathrm{Nn} \pi \mathrm{b}^2 2 \mathrm{mt}
\end{aligned}
$$
Thus, current varies as a function of $\left(\mathrm{mt}^2+\mathrm{C}\right)$. So, net emf produced in $\mathrm{N}$ turns of smaller coil
$$
e=-\mu_0 N n \pi b^2 2 m t
$$
So, magnetic field due to a solenoid $S_1$ is
$$
\mathrm{B}=\mu_0 \mathrm{n} l
$$
Magnetic flux in smaller coil
$$
\begin{aligned}
\phi &=\mathrm{NBA} \\
\mathrm{A} &=\pi \mathrm{b}^2
\end{aligned}
$$
By applying Faraday's law of EMI, the induced emf in coil due to solenoids verying magnetic field.
$$
e=\frac{-d \phi}{d t} \quad(\because \phi=N B A)
$$
So, $e=\frac{-d}{d t}(\mathrm{NBA}) \quad\left(\therefore \mathrm{B}=\mu_0 \mathrm{Ni}\right)$
$$
\begin{aligned}
\mathrm{e}=-\mathrm{N} \pi \mathrm{b}^2 \mu_0 \mathrm{n} \frac{\mathrm{dl}}{\mathrm{dt}} &=-\mathrm{Nn} \pi \mu_0 \mathrm{~b}^2 \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{mt}^2+\mathrm{C}\right) \\
&=-\mu \mathrm{Nn} \pi \mathrm{b}^2 2 \mathrm{mt}
\end{aligned}
$$
Thus, current varies as a function of $\left(\mathrm{mt}^2+\mathrm{C}\right)$. So, net emf produced in $\mathrm{N}$ turns of smaller coil
$$
e=-\mu_0 N n \pi b^2 2 m t
$$
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