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A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}$ Wb. The self-inductance of the solenoid is -
A
1.0 henry
B
4.0 henry
C
2.5 henry
D
2.0 henry
Detailed Solution
Total flux linkage = flux through each turn $\times$ number of turns.
$N\phi = 500 \times 4 \times 10^{-3} = 2$ Wb
By definition of self-inductance, total flux linkage $N\phi = LI$.
$\Rightarrow L = \dfrac{N\phi}{I} = \dfrac{2}{2} = 1$ H
So the self-inductance of the solenoid is 1.0 henry.
$N\phi = 500 \times 4 \times 10^{-3} = 2$ Wb
By definition of self-inductance, total flux linkage $N\phi = LI$.
$\Rightarrow L = \dfrac{N\phi}{I} = \dfrac{2}{2} = 1$ H
So the self-inductance of the solenoid is 1.0 henry.