KCET · Physics · Electromagnetic Induction
The current in a coil of inductance \(0.2 \mathrm{H}\) changes from \(5 \mathrm{~A}\) to \(2 \mathrm{~A}\) in \(0.5 \mathrm{~s}\). The magnitude of the average induced emf in the coil is
- A \(0.6 \mathrm{~V}\)
- B \(1.2 \mathrm{~V}\)
- C \(30 \mathrm{~V}\)
- D \(0.3 \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(1.2 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Given, \(L=0.2 \mathrm{H}, I_{1}=5 \mathrm{~A}, I_{2}=2 \mathrm{~A}\) and \(\Delta t=0.5 \mathrm{~s}\)
The magnitude of induced emf, \(|e|=L \frac{d I}{d t}\)
\(=L \frac{\Delta I}{\Delta t}=\frac{0.2 \times(5-2)}{0.5}=1.2 \mathrm{~V}\)
The magnitude of induced emf, \(|e|=L \frac{d I}{d t}\)
\(=L \frac{\Delta I}{\Delta t}=\frac{0.2 \times(5-2)}{0.5}=1.2 \mathrm{~V}\)
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