COMEDK · Physics · 22. Electromagnetic Induction
The current in a coil changes from \(1 \mathrm{~mA}\) to \(5 \mathrm{~mA}\) in \(4 \mathrm{milli}\) secon(d) If the coefficient of self-induction of the coil is \(10 \mathrm{mH}\), then the magnitude of the self-induced emf is
- A \(10 \mathrm{mV}\)
- B \(5 \mathrm{mV}\)
- C \(2.5 \mathrm{mV}\)
- D \(1 \mathrm{mV}\)
Answer & Solution
Correct Answer
(A) \(10 \mathrm{mV}\)
Step-by-step Solution
Detailed explanation
Given, \(\quad I_{1}=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}\) \(I_{2}=5 \mathrm{~mA}=5 \times 10^{-3} \mathrm{~A}\) \(t=4 \mathrm{~ms}=4 \times 10^{-3} \mathrm{~s}\) \(L=10 \mathrm{~m} \mathrm{H}=10 \times 10^{-3} \mathrm{H}\) \(\therefore\) Self-induced emf,…
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