A current of 0.5 A is passed through winding of a long solenoid having 400 turns. The magnetic flux linked…

A current of 0.5 A is passed through winding of a long solenoid having 400 turns. The magnetic flux linked with each turn is $3 \times 10^{-3} \mathrm{~Wb}$. The self inductance of the solenoid is
  1. 2.4 H
  2. 2.0 H
  3. 1.2 H
  4. 0.6 H

Solution

Self inductance, $L=\frac{N \phi}{i}$ $\therefore \quad \mathrm{L}=\frac{400 \times 3 \times 10^{-3}}{0.5}=\frac{1200 \times 10^{-3}}{0.5}=2.4 \mathrm{H}$

Asked in: MHT CET 2024 (04 May Shift 2)

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