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A long solenoid has 1000 turns. When a current of 4 A flows through it, the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}$ Wb. The self-inductance of the solenoid is
A
3 H
B
2 H
C
1 H
D
4 H
Explanation
$L = N\phi/I$.
Detailed Solution
Total flux linked with the solenoid $= N\phi = 1000\times4\times10^{-3}$ Wb = 4 Wb
Self-inductance: $N\phi = LI \Rightarrow 4 = L\times4$
L = 1 H
Self-inductance: $N\phi = LI \Rightarrow 4 = L\times4$
L = 1 H