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A coil of wire of radius $r$ has 600 turns and self inductance of $108 \mathrm{mH}$. The self inductance of a coil with same radius and 500 turns is
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1218 Upvotes
Verified Answer
The correct answer is:
90 mH
Self-inductance of a coil $=\frac{N \phi_B}{I}$
$$
=\frac{N \frac{\mu_0 I}{2 R} \times \pi R^2}{I}=\frac{\pi \mu_0}{2} \cdot N R
$$
So,
$$
\begin{aligned}
& \frac{L_2}{L_1}=\frac{N_2}{N_1} \\
& L_2=L_1 \times \frac{N_2}{N_1}=108 \times \frac{500}{600}=90 \mathrm{mH}
\end{aligned}
$$
$$
=\frac{N \frac{\mu_0 I}{2 R} \times \pi R^2}{I}=\frac{\pi \mu_0}{2} \cdot N R
$$
So,
$$
\begin{aligned}
& \frac{L_2}{L_1}=\frac{N_2}{N_1} \\
& L_2=L_1 \times \frac{N_2}{N_1}=108 \times \frac{500}{600}=90 \mathrm{mH}
\end{aligned}
$$
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