AP EAMCET · PHYSICS · Electromagnetic Induction
The current in an inductor of self-inductance \(L=40 \mathrm{mH}\) is to be increased uniformly from \(2 \mathrm{~A}\) to \(12 \mathrm{~A}\) in \(8 \mathrm{~ms}\). The emf induced in the inductor during this process is
- A \(50 \mathrm{~V}\)
- B \(0.4 \mathrm{~V}\)
- C \(40 \mathrm{~V}\)
- D \(100 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(50 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, Self-inductance of inductor, \(L=40 \mathrm{mH}\) Initial current, \(I_1=2 \mathrm{~A}\) Final current, \(I_2=12 \mathrm{~A}\) Time interval, \(d t=8 \mathrm{~ms}\) Using expression for induced emf, \(\varepsilon=-L \frac{d i}{d t}\)…
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