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A coil of wire of radius $r$ has 600 turns and a self inductance of $108 \mathrm{mH}$. The self inductance of a coil with same radius and 500 turns is
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Verified Answer
The correct answer is:
$75 \mathrm{mH}$
The given,
$L_1=108 \mathrm{mH}, N_1=600$ turns, $N_2=500$ turns and $L_2=$ ?
By self-inductance of a plane coil
$$
\begin{aligned}
& L_1=\frac{\mu_0 \pi N_1^2 a_1}{2} \\
& L_2=\frac{\mu_0 \pi N_2^2 a_2}{2}
\end{aligned}
$$
From the Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{L_1}{L_2} & =\left(\frac{N_1}{N_2}\right)^2 \quad\left(\because a_1=a_2\right) \\
\frac{108}{L_2} & =\left(\frac{600}{500}\right)^2 \\
L_2 & =\frac{108 \times 25}{36} \\
L_2 & =3 \times 25=75 \mathrm{mH}
\end{aligned}
$$
$L_1=108 \mathrm{mH}, N_1=600$ turns, $N_2=500$ turns and $L_2=$ ?
By self-inductance of a plane coil
$$
\begin{aligned}
& L_1=\frac{\mu_0 \pi N_1^2 a_1}{2} \\
& L_2=\frac{\mu_0 \pi N_2^2 a_2}{2}
\end{aligned}
$$
From the Eqs. (i) and (ii), we get
$$
\begin{aligned}
\frac{L_1}{L_2} & =\left(\frac{N_1}{N_2}\right)^2 \quad\left(\because a_1=a_2\right) \\
\frac{108}{L_2} & =\left(\frac{600}{500}\right)^2 \\
L_2 & =\frac{108 \times 25}{36} \\
L_2 & =3 \times 25=75 \mathrm{mH}
\end{aligned}
$$
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