33. 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 $4 \times 10^{-3} Tm^2$. The self inductance of the solenoid is: (2008)
Solution:
Total flux linked is given by $N \Phi = L I$.
Here, $N = 500$, $\Phi = 4 \times 10^{-3} Wb$ (or $T m^2$), and $I = 2 A$.
$L = \frac{N \Phi}{I} = \frac{500 \times 4 \times 10^{-3}}{2} = 1.0 henry$.
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