A 800 turn coil of effective area 0.05 m^2 is kept perpendicular to a magnetic field 5 × 10^-5 T . When the plane of the coil is rotated by 90^ around - Physics MCQ NEET 2019 2026 | ExamTest.live

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A 800 turn coil of effective area 0.05 m^{2} is kept perpendicular to a magnetic field 5 \times 1 0^{- 5} T . When the plane of the coil is rotated by 9 0^{\circ} around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:

Solution & Step-by-step Explanation

Induced emf is given by Faraday's Law: ∣ ϵ ∣ = N \frac{Δ ϕ}{Δ t} Initial flux ϕ_{i} = B A cos 0^{\circ} (perpendicular to field means θ between normal and field is 0^{\circ}). ϕ_{i} = (5 \times 1 0^{- 5}) (0.05) = 2.5 \times 1 0^{- 6} Wb .Final flux ϕ_{f} = B A cos 9 0^{\circ} = 0 . ϵ = 800 \times \frac{2.5 \times 1 0 ^{- 6} - 0}{0.1} = 800 \times 2.5 \times 1 0^{- 5} = 2000 \times 1 0^{- 5} = 0.02 V

Try it yourself before checking the explanation above.

A 800 turn coil of effective area 0.05 m^{2} is kept perpendicular to a magnetic field 5 \times 1 0^{- 5} T . When the plane of the coil is rotated by 9 0^{\circ} around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:

2 V

0.2 V

2 \times 1 0^{- 3} V

0.02 V

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