MHT CET · Physics · Electromagnetic Induction
Two concentric circular coils having radii \(r_{1}\) and \(r_{2},\left(r_{2}< < r_{1}\right)\) are placed co-
axially with centres coinciding. The mutual induction of the arrangement is
(Both coils have single turn) \(\left(\mu_{0}=\right.\) permeability of free space)
- A \(\frac{\mu_{0} \pi r_{2}^{2}}{r_{1}}\)
- B \(\frac{\mu_{0} \pi r_{1}^{2}}{r_{2}}\)
- C \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\)
- D \(\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}\)
Step-by-step Solution
Detailed explanation
If current \(\mathrm{I}_{1}\) flows in the coil of radius \(\mathrm{r}_{1}\), the magnetic field at the centre is given by
\(\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{r}_{1}}\)
The magnetic flux passing through the coil of radius \(\mathrm{r}_{2}\) will be
\(\phi_{2}=\mathrm{B}_{1} \times \pi \mathrm{r}_{2}^{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{r}_{1}} \times \pi \mathrm{r}_{2}^{2}\)
Mutual inductance \(\quad M=\frac{\phi_{2}}{I_{1}}=\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}\)
\(\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{2 \mathrm{r}_{1}}\)
The magnetic flux passing through the coil of radius \(\mathrm{r}_{2}\) will be
\(\phi_{2}=\mathrm{B}_{1} \times \pi \mathrm{r}_{2}^{2}=\frac{\mu_{\mathrm{o}} \mathrm{I}_{1}}{2 \mathrm{r}_{1}} \times \pi \mathrm{r}_{2}^{2}\)
Mutual inductance \(\quad M=\frac{\phi_{2}}{I_{1}}=\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}\)
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