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A circular coil with a cross-sectional area of $4 \mathrm{~cm}^2$ has 10 turns. It is placed at the center of a long solenoid that has 15 turns $/ \mathrm{cm}$ and a crosssectional area of $10 \mathrm{~cm}^2$, as shown in the figure. The axis of the coil conicides with the axis of the solenoid. What is their mutual inductance?
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1465 Upvotes
Verified Answer
The correct answer is:
$7.54 \mu \mathrm{H}$
Let us refer to the coil as circuit 1 and the solenoid as circuit 2 . The field in the central region of the solenoid is uniform, so the flux through the coil is
$$
\Phi_{12}=B_2 A_1=\left(\mu_0 n_2 I_2\right) A_1
$$
where $n_2=N_2 / l=1500$ turns $/ \mathrm{m}$. The mutual inductance is
$$
\begin{aligned}
& M=\frac{N_1 \Phi_{12}}{I_2}=\mu_0 n_2 N_1 A_1 \\
& =\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(1500 \mathrm{~m}^{-1}\right)(10)\left(4 \times 10^{-4} \mathrm{~m}^2\right) \\
& \quad=7.54 \times 10^{-6} \mathrm{H}
\end{aligned}
$$
$$
\Phi_{12}=B_2 A_1=\left(\mu_0 n_2 I_2\right) A_1
$$
where $n_2=N_2 / l=1500$ turns $/ \mathrm{m}$. The mutual inductance is
$$
\begin{aligned}
& M=\frac{N_1 \Phi_{12}}{I_2}=\mu_0 n_2 N_1 A_1 \\
& =\left(4 \pi \times 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}\right)\left(1500 \mathrm{~m}^{-1}\right)(10)\left(4 \times 10^{-4} \mathrm{~m}^2\right) \\
& \quad=7.54 \times 10^{-6} \mathrm{H}
\end{aligned}
$$
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