MHT CET · Physics · Electromagnetic Induction
Two concentric circular coils having radii \(r_1\) and \(\mathrm{r}_2\left(\mathrm{r}_2 \ll \mathrm{r}_1\right)\) are placed co-axially with centres coinciding. The mutual induction of the arrangement is (Both coils have single turn, \(\mu_0=\) permeability of free space)
- A \(\frac{\mu_0 \pi r_2^2}{2 r_1}\)
- B \(\frac{\mu_0 \pi r_2}{2 r_1}\)
- C \(\frac{\mu_0 \pi r_2^2}{r_1^2}\)
- D \(\frac{\mu_0 \pi r_2}{r_1}\)
Answer & Solution
Correct Answer
(A) \(\frac{\mu_0 \pi r_2^2}{2 r_1}\)
Step-by-step Solution
Detailed explanation
Let \(I_1\) be the current through the coil whose radius is \(r_1\).
\(\therefore \quad\) Magnetic field at the centre of the coil,
\(B_1=\frac{\mu_0 I_1}{2 r_1}\)
Magnetic flux passing through the coil of radius \(\mathrm{r}_2\) is
\(\begin{aligned}
& \phi_2=\mathrm{B}_1 \cdot \pi \mathrm{r}_2^2 \quad(\because \phi=\mathrm{B} . \mathrm{A}) \\
& =\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{r}_1} \cdot \pi \mathrm{r}_2^2
\end{aligned}\)
\(\therefore \quad\) Mutual inductance of the arrangement,
\(\begin{aligned}
\mathbf{M} & =\frac{\phi_2}{\mathrm{I}_1} \\
& =\frac{\mu_0 \mathrm{I}_1 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1 \mathrm{I}_1} \\
& =\frac{\mu_0 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1}
\end{aligned}\)
\(\therefore \quad\) Magnetic field at the centre of the coil,
\(B_1=\frac{\mu_0 I_1}{2 r_1}\)
Magnetic flux passing through the coil of radius \(\mathrm{r}_2\) is
\(\begin{aligned}
& \phi_2=\mathrm{B}_1 \cdot \pi \mathrm{r}_2^2 \quad(\because \phi=\mathrm{B} . \mathrm{A}) \\
& =\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{r}_1} \cdot \pi \mathrm{r}_2^2
\end{aligned}\)
\(\therefore \quad\) Mutual inductance of the arrangement,
\(\begin{aligned}
\mathbf{M} & =\frac{\phi_2}{\mathrm{I}_1} \\
& =\frac{\mu_0 \mathrm{I}_1 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1 \mathrm{I}_1} \\
& =\frac{\mu_0 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1}
\end{aligned}\)
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