KCET · Physics · Electromagnetic Induction
A long solenoid has 500 turns, when a current of \(2 \mathrm{~A}\) is passed through it, the resulting magnetic flux linked with each turn of the solenoid is \(4 \times 10^{-3} \mathrm{~Wb}\), then selfinduction of the solenoid is
- A \(2.5 \mathrm{H}\)
- B \(2.0 \mathrm{H}\)
- C \(1.0 \mathrm{H}\)
- D \(4.0 \mathrm{H}\)
Answer & Solution
Correct Answer
(C) \(1.0 \mathrm{H}\)
Step-by-step Solution
Detailed explanation
Given, number of turns in the solenoid,
\(N=500\)
Current, \(I=2 \mathrm{~A}\)
Magnetic flux, \(\phi=4 \times 10^{-3} \mathrm{~Wb}\)
Self-inductance of coil,
\(L=\frac{N \phi}{I}=\frac{500 \times 4 \times 10^{-3}}{2}=1 \mathrm{H}\)
\(N=500\)
Current, \(I=2 \mathrm{~A}\)
Magnetic flux, \(\phi=4 \times 10^{-3} \mathrm{~Wb}\)
Self-inductance of coil,
\(L=\frac{N \phi}{I}=\frac{500 \times 4 \times 10^{-3}}{2}=1 \mathrm{H}\)
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