A coil has an area 0.06 m² and it has 600 turns. After placing the coil in a magnetic field of strength 5 × 10⁻⁵ Wb m⁻², it is rotated through 90° in 0.2 second. The magnitude of average e.m.f induced in the coil is [cos 0° = sin 90° = 1 and sin 0° = cos 90° = 0] (a) 12 × 10⁻³ V (b) 3 mV (c) 3 V (d) 9 × 10⁻³ V

Answer: (d). Rotating through 90° from the perpendicular position: flux goes from NBA to 0, so Δφ = NBA. NBA = 600 × 5 × 10⁻⁵ × 0.06 = 1.8 × 10⁻³ Wb e = 1.8 × 10⁻³0.2 = 9 × 10⁻³ V So the answer is (d) 9 × 10⁻³ V.

A coil has an area 0.06 m² and it has 600 turns. After placing the… | MHT CET 2021 (24 Sep, Shift 1)

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Magnetic Flux & Faraday's Law

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

$[\cos 0^\circ = \sin 90^\circ = 1$ and $\sin 0^\circ = \cos 90^\circ = 0]$

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Answer: (d)

Rotating through $90^\circ$ from the perpendicular position: flux goes from $NBA$ to $0$ , so $\Delta\phi = NBA$ .
$NBA = 600 \times 5 \times 10^{-5} \times 0.06 = 1.8 \times 10^{-3}$ Wb
$e = \frac{1.8 \times 10^{-3}}{0.2} = 9 \times 10^{-3}$ V
So the answer is (d) $9 \times 10^{-3}$ V.

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Magnetic Flux & Faraday's Law question from MHT CET 2021 (24 Sep, Shift 1)

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