Question:
When a current in the conducting coil is changed from 5 A in one direction to 5 A in opposite direction in 0.5 second, an average induced e.m.f in the coil is 2 V . The selfinductance of the coil is
When a current in the conducting coil is changed from 5 A in one direction to 5 A in opposite direction in 0.5 second, an average induced e.m.f in the coil is 2 V . The selfinductance of the coil is
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Use the average emf as the self inductance times the rate of change of current, noting the current reverses direction.
Updated On: Nov 13, 2026
- 25 mH
- 50 mH
- 75 mH
- 100 mH
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
The average induced emf in a coil is \(e=L\left|\dfrac{\Delta I}{\Delta t}\right|\).
Step 2: Change in current
The current changes from \(+5\ \text{A}\) to \(-5\ \text{A}\), so \(\Delta I=10\ \text{A}\) in \(\Delta t=0.5\ \text{s}\).\n\(\dfrac{\Delta I}{\Delta t}=\dfrac{10}{0.5}=20\ \text{A/s}\)
Step 3: Calculation
\(L=\dfrac{e}{\Delta I/\Delta t}=\dfrac{2}{20}=0.1\ \text{H}=100\ \text{mH}\)
Final Answer:
The self inductance of the coil is 100 mH. \[ \boxed{\text{(D) }\text{100 mH}} \]
The average induced emf in a coil is \(e=L\left|\dfrac{\Delta I}{\Delta t}\right|\).
Step 2: Change in current
The current changes from \(+5\ \text{A}\) to \(-5\ \text{A}\), so \(\Delta I=10\ \text{A}\) in \(\Delta t=0.5\ \text{s}\).\n\(\dfrac{\Delta I}{\Delta t}=\dfrac{10}{0.5}=20\ \text{A/s}\)
Step 3: Calculation
\(L=\dfrac{e}{\Delta I/\Delta t}=\dfrac{2}{20}=0.1\ \text{H}=100\ \text{mH}\)
Final Answer:
The self inductance of the coil is 100 mH. \[ \boxed{\text{(D) }\text{100 mH}} \]
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