MHT CET · Physics · Electromagnetic Induction
Two coils A and B have mutual inductance 0.008 \(\mathrm{H}\). The current changes in the coil \(\mathrm{A}\), according to the equation \(\mathrm{I}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}\), where \(I_m=5 A\) and \(\omega=200 \pi \mathrm{rad} \mathrm{s}^{-1}\). The maximum value of the e.m.f. induced in the coil B in volt is
- A \(4 \pi\)
- B \(8 \pi\)
- C \(10 \pi\)
- D \(16 \pi\)
Answer & Solution
Correct Answer
(B) \(8 \pi\)
Step-by-step Solution
Detailed explanation
\(\mathrm{e}=\mathrm{M} \frac{\mathrm{dI}}{\mathrm{dt}}\)
Given: \(\mathrm{M}=0.008, \mathrm{I}_{\mathrm{m}}=5 \mathrm{~A}, \omega=200 \mathrm{rad} / \mathrm{s}\) \(\therefore \quad \mathrm{e}=0.008 \times \mathrm{I}_{\mathrm{m}} \omega \mathrm{e} \mathrm{os} \omega \mathrm{t}\)
For \(\mathrm{e}=\mathrm{e}_{\max }, \cos \omega \mathrm{t}=1\)
\(\begin{aligned}
\therefore \quad \mathrm{e}_{\max } & =0.008 \times \mathrm{I}_{\mathrm{m}} \times \omega \\
& =0.008 \times 5 \times 200 \pi \\
& =8 \pi
\end{aligned}\)
Given: \(\mathrm{M}=0.008, \mathrm{I}_{\mathrm{m}}=5 \mathrm{~A}, \omega=200 \mathrm{rad} / \mathrm{s}\) \(\therefore \quad \mathrm{e}=0.008 \times \mathrm{I}_{\mathrm{m}} \omega \mathrm{e} \mathrm{os} \omega \mathrm{t}\)
For \(\mathrm{e}=\mathrm{e}_{\max }, \cos \omega \mathrm{t}=1\)
\(\begin{aligned}
\therefore \quad \mathrm{e}_{\max } & =0.008 \times \mathrm{I}_{\mathrm{m}} \times \omega \\
& =0.008 \times 5 \times 200 \pi \\
& =8 \pi
\end{aligned}\)
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