A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux…

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 $4 \times 10^{-3} \omega \mathrm{b}$. The self inductance of the solenoid is
  1. 4.0 henry
  2. 2.5 henry
  3. 2.0 henry
  4. 1.0 henry

Solution

$\begin{aligned} & N \phi=L i \\ & 500 \times 4 \times 10^{-3}=2 L \\ & L=1.0 \text { henry } \end{aligned}$

Asked in: NEET 2008 (Mains)

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