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A long solenoid with 15 turns per $\mathrm{cm}$ has a small loop of area $2.0 \mathrm{~cm}^2$ placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from $2.0 \mathrm{~A}$ to $4.0 \mathrm{~A}$ in $0.1 \mathrm{~s}$, what is the induced emf in the loon while the current is changing?
Solution:
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Verified Answer
Given: no. of turns $=15 / \mathrm{cm}$
area $=2 \mathrm{~cm}^2$
change in current $\mathrm{dI}=4-2=2 \mathrm{~A}$
change in time $\mathrm{dt}=0.1 \mathrm{~s}$
To find: Induced emf=e
Formula used: $e=-\frac{d \phi}{d t}=\frac{-d}{d t}\left(\mu_0 n\right.$ IA $)$
Magnetic flux $\phi=\mathrm{BA}=\mu_0 \mathrm{n}$ IA
induced emf $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\mu_0 \mathrm{nA} \frac{\mathrm{dI}}{\mathrm{dt}}$
$$
\begin{aligned}
&\mu_0=4 \pi \times 10^{-7}, \mathrm{n}=15 \mathrm{~cm}^{-1}=1500 \mathrm{~m}^{-1} \\
&\mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
$e=-4 \pi \times 10^{-7} \times 1500 \times 2 \times 10^{-4} \times \frac{4-2}{0.1}$
$$
\begin{aligned}
&=-4 \pi \times 3 \times 10^{-7-4+3} \times 20 \\
&=-4 \times 3.14 \times 6 \times 10^{-7}=-7.54 \times 10^{-6} \mathrm{~V}
\end{aligned}
$$
area $=2 \mathrm{~cm}^2$
change in current $\mathrm{dI}=4-2=2 \mathrm{~A}$
change in time $\mathrm{dt}=0.1 \mathrm{~s}$
To find: Induced emf=e
Formula used: $e=-\frac{d \phi}{d t}=\frac{-d}{d t}\left(\mu_0 n\right.$ IA $)$
Magnetic flux $\phi=\mathrm{BA}=\mu_0 \mathrm{n}$ IA
induced emf $\mathrm{e}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=-\mu_0 \mathrm{nA} \frac{\mathrm{dI}}{\mathrm{dt}}$
$$
\begin{aligned}
&\mu_0=4 \pi \times 10^{-7}, \mathrm{n}=15 \mathrm{~cm}^{-1}=1500 \mathrm{~m}^{-1} \\
&\mathrm{~A}=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$
$e=-4 \pi \times 10^{-7} \times 1500 \times 2 \times 10^{-4} \times \frac{4-2}{0.1}$
$$
\begin{aligned}
&=-4 \pi \times 3 \times 10^{-7-4+3} \times 20 \\
&=-4 \times 3.14 \times 6 \times 10^{-7}=-7.54 \times 10^{-6} \mathrm{~V}
\end{aligned}
$$
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