MHT CET · Physics · Electromagnetic Induction
A circular coil of radius ' \(\mathrm{R}\) ' has ' \(\mathrm{N}\) ' turns of a wire. The coefficient of self-induction of the coil will be ( \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 N \pi R^2}{2}\)
- B \(\frac{\mu_0 \mathrm{~N} \pi \mathrm{R}}{4}\)
- C \(\frac{\mu_0 N^2 \pi R}{2}\)
- D \(\frac{\mu_0 \mathrm{~N} \pi \mathrm{R}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mu_0 N^2 \pi R}{2}\)
Step-by-step Solution
Detailed explanation
Magnetic field at the center of the coil is given by
\(
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
\)
Magnetic flux linked to the coil is
\(
\begin{aligned}
& \phi=\mathrm{NBA}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{I}}{2 \mathrm{R}} \cdot \pi \mathrm{R}^2=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{RI}}{2} \\
& \mathrm{~L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{R}}{2}
\end{aligned}
\)
\(
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
\)
Magnetic flux linked to the coil is
\(
\begin{aligned}
& \phi=\mathrm{NBA}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{I}}{2 \mathrm{R}} \cdot \pi \mathrm{R}^2=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{RI}}{2} \\
& \mathrm{~L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{R}}{2}
\end{aligned}
\)
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