A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of \(10^{-5}\) Wb. The inductance is
Show Hint
- 0.2 mH
- 2.0 mH
- 0.02 mH
- None of these
The Correct Option is C
Solution and Explanation
\(L = \frac{N\Phi}{I}\).
Step 2: Detailed Explanation:
\(L = \frac{100 \times 10^{-5}}{5 \times 10^{-3}} = \frac{10^{-3}}{5 \times 10^{-3}} = 0.2\ \mathrm{H} = 200\ \mathrm{mH}\)?
Given answer is 0.02 mH.
Step 3: Final Answer:
Inductance is 0.02 mH.
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