KCET · Physics · Electromagnetic Induction
The current in a coil changes from \(2 \mathrm{~A}\) to \(5 \mathrm{~A}\) in \(0.3 \mathrm{~s}\). The magnitude of emf induced in the coil is \(1.0 \mathrm{~V}\). The value of self-inductance of the coil is
- A \(1.0 \mathrm{mH}\)
- B \(100 \mathrm{mH}\)
- C \(0.1 \mathrm{mH}\)
- D \(10 \mathrm{mH}\)
Answer & Solution
Correct Answer
(B) \(100 \mathrm{mH}\)
Step-by-step Solution
Detailed explanation
\(\Delta I=5-2=3 A\)
\(\Delta t=0.3 \mathrm{~s}\)
Induced emf, \(e=1 \mathrm{~V}\)
We know that, \(e=\frac{L d I}{d t}\)
\(\Rightarrow e=\frac{L \Delta I}{\Delta t}\)
\(\Rightarrow L =\frac{e}{\frac{\Delta t}{\Delta t}}=\frac{1}{\frac{3}{0.3}}=0.1 \mathrm{H}=100 \mathrm{mH}\)
\(\Delta t=0.3 \mathrm{~s}\)
Induced emf, \(e=1 \mathrm{~V}\)
We know that, \(e=\frac{L d I}{d t}\)
\(\Rightarrow e=\frac{L \Delta I}{\Delta t}\)
\(\Rightarrow L =\frac{e}{\frac{\Delta t}{\Delta t}}=\frac{1}{\frac{3}{0.3}}=0.1 \mathrm{H}=100 \mathrm{mH}\)
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