MHT CET · Physics · Electromagnetic Induction
Two coils \(\mathrm{P}\) and S have a mutual inductance of \(3 \times 10^{-3} \mathrm{H}\). If the current in the coil, P is \(I=20 \sin (50 \pi t) \mathrm{A}\), then the maximum value of the e.m.f. induced in coil \(\mathrm{S}\) is
- A \(15.70 \mathrm{~V}\)
- B \(9.42 \mathrm{~V}\)
- C \(3.14 \mathrm{~V}\)
- D \(6.25 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(9.42 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The correct option is (B).
Concept: Flux associated among the coils is \(\phi=\mathrm{MI}\) and the induced emf is given by \(\mathrm{E}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\).
Therefore, \(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\),
Therefore, \(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\), therefore
\(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\), therefore
\(E=\left(-3 \times 10^{-3} \mathrm{H} \times 50 \pi \times 20\right) \cos (50 \pi \mathrm{t}) \mathrm{A}\)
The maximum emf is given by
\(\left|E_0\right|=3 \times 10^{-3} \times 50 \pi \times 20 \mathrm{~V}=9.42 \mathrm{~V}\)
Concept: Flux associated among the coils is \(\phi=\mathrm{MI}\) and the induced emf is given by \(\mathrm{E}=-\frac{\mathrm{d} \phi}{\mathrm{dt}}\).
Therefore, \(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\),
Therefore, \(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\), therefore
\(\mathrm{E}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}\).
Given, \(\mathrm{I}=20 \sin (50 \pi \mathrm{t})\) and \(\mathrm{M}=3 \times 10^{-3} \mathrm{H}\), therefore
\(E=\left(-3 \times 10^{-3} \mathrm{H} \times 50 \pi \times 20\right) \cos (50 \pi \mathrm{t}) \mathrm{A}\)
The maximum emf is given by
\(\left|E_0\right|=3 \times 10^{-3} \times 50 \pi \times 20 \mathrm{~V}=9.42 \mathrm{~V}\)
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