MHT CET · Physics · Electromagnetic Induction
When a current of \(4 \mathrm{~A}\) changes to \(8 \mathrm{~A}\) in \(0.6 \mathrm{~s}\) in a primary coil it induces an e. m. f. of \(50 \mathrm{mV}\) in secondary coil. The mutual inductance between two coils is
- A \(1.2 \mathrm{mH}\)
- B \(3.33 \mathrm{mH}\)
- C \(7.5 \mathrm{mH}\)
- D \(10.5 \mathrm{mH}\)
Answer & Solution
Correct Answer
(C) \(7.5 \mathrm{mH}\)
Step-by-step Solution
Detailed explanation
Change in current in primary coil,
\(\mathrm{di}_1=(8-4) \mathrm{A}=4 \mathrm{~A}\)
Time taken, \(\mathrm{dt}=0.6 \mathrm{~s}\)
Induced e. m. f in secondary coil,
\(\mathrm{E}_2=50 \times 10^{-3} \mathrm{~V}\)
So, mutual inductance is,
\(M=\frac{E_2}{\mathrm{di}_1 / \mathrm{dt}}=\frac{\mathrm{E}_2 \mathrm{dt}}{\mathrm{di}_1}=\frac{50 \times 10^{-3} \mathrm{~V} \times 0.6 \mathrm{~s}}{4 \mathrm{~A}}=\) \(7.5 \times 10^{-3} \mathrm{H}=7.5 \mathrm{mH}\)
\(\mathrm{di}_1=(8-4) \mathrm{A}=4 \mathrm{~A}\)
Time taken, \(\mathrm{dt}=0.6 \mathrm{~s}\)
Induced e. m. f in secondary coil,
\(\mathrm{E}_2=50 \times 10^{-3} \mathrm{~V}\)
So, mutual inductance is,
\(M=\frac{E_2}{\mathrm{di}_1 / \mathrm{dt}}=\frac{\mathrm{E}_2 \mathrm{dt}}{\mathrm{di}_1}=\frac{50 \times 10^{-3} \mathrm{~V} \times 0.6 \mathrm{~s}}{4 \mathrm{~A}}=\) \(7.5 \times 10^{-3} \mathrm{H}=7.5 \mathrm{mH}\)
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