Q114 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

When a current of $1 \mathrm{~A}$ is passed through a coil of 100 turns, the flux associated with it is $2.5 \times 10^{-5} \mathrm{~Wb} /$ turn. The self inductance of the coil in millihenry is

  • A
    $40$
  • B
    $25$
  • C
    $4$
  • Correct answer
    D
    $2.5$

Solution

Given: $N=100, I=1 \mathrm{~A}$, $\phi=2.5 \times 10^{-5} \mathrm{~Wb} /$ turn Self-inductance of coil, $$ \begin{aligned} \mathrm{L} & =\frac{\mathrm{N} \phi}{\mathrm{I}} \\ \therefore \quad \mathrm{L} & =\frac{100 \times 2.5 \times 10^{-5}}{1}=2.5 \times 10^{-3} \mathrm{H}=2.5 \mathrm{mH} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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