MHT CET · Physics · Electromagnetic Induction
Two coaxial coils A and B of radii 'R ' and 'Rz' are placed in the same plane. (R \(_{2}>\)
\(\mathrm{R}_{1}\) ). If a current is passed through coil \(\mathrm{B}\), the coefficient of mutual inductance
between the coils is proportional to
- A \(\frac{1}{\mathrm{R}_{1} \mathrm{R}_{2}}\)
- B \(\frac{\mathrm{R}_{2}^{2}}{\mathrm{R}_{1}}\)
- C \(R_{1} R_{2}\)
- D \(\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}\)
Answer & Solution
Correct Answer
(D) \(\frac{\mathrm{R}_{1}^{2}}{\mathrm{R}_{2}}\)
Step-by-step Solution
Detailed explanation
Magnetic field at the centre of primary coil
\(
B=\mu_{0} i_{1} / 2 R_{1} .
\)
Considering it to be uniform, magnetic flux passing through secondary coil is
\(
\phi_{2}=B A=\frac{\mu_{0} i_{1}}{2 R_{1}}\left(\pi R_{2}^{2}\right)
\)
Now, \(M=\frac{\phi_{2}}{i_{1}}=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}\)
\(
\therefore M \propto \frac{R_{2}^{2}}{R_{1}} .
\)
\(
B=\mu_{0} i_{1} / 2 R_{1} .
\)
Considering it to be uniform, magnetic flux passing through secondary coil is
\(
\phi_{2}=B A=\frac{\mu_{0} i_{1}}{2 R_{1}}\left(\pi R_{2}^{2}\right)
\)
Now, \(M=\frac{\phi_{2}}{i_{1}}=\frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}\)
\(
\therefore M \propto \frac{R_{2}^{2}}{R_{1}} .
\)
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