Q126 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

A coil has an area $0.06 \mathrm{~m}^2$ and it has 600 turns. After placing the coil in a magnetic field of strength $5 \times 10^{-5} \mathrm{Wbm}^{-2}$, it is rotated through $90^{\circ}$ in 0.2 second. The magnitude of average e.m.f induced in the coil is

  • A
    $12 \times 10^{-3} \mathrm{~V}$
  • B
    $3 \mathrm{mV}$
  • C
    $3 \mathrm{~V}$
  • Correct answer
    D
    $9 \times 10^{-3} \mathrm{~V}$

Solution

Induced emf, $\mathrm{e}=\frac{-\mathrm{N}\left(\phi_2-\phi_1\right)}{\mathrm{t}}=\frac{-\mathrm{NBA}\left(\cos \theta_2-\cos \theta_1\right)}{\mathrm{t}}$ Assuming that the coil was initially placed perpendicular magnetic flux $$ \begin{aligned} & \theta_1=0^{\circ} \text { and } \theta_2=90^{\circ} . \\ & \mathrm{e}=\frac{-\mathrm{NBA}\left(\cos 90^{\circ}-\cos 0^{\circ}\right)}{\mathrm{t}} \\ & =\frac{-\mathrm{NBA}(0-1)}{\mathrm{t}}=\frac{\mathrm{NBA}}{\mathrm{t}} \\ & =\frac{600 \times 5 \times 10^{-5} \times 0.06}{0.2}=9 \times 10^{-3} \mathrm{~V} \end{aligned} $$

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

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