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NEET PHYSICSMedium

A 800 turn coil of effective area 0.05 m20.05 \text{ m}^2 is kept perpendicular to a magnetic field 5×10−5 T5 \times 10^{-5} \text{ T}. When the plane of the coil is rotated by 90∘90^\circ around any of its coplanar axis in 0.1 s0.1 \text{ s}, the emf induced in the coil will be:

A

2 V2 \text{ V}

B

0.2 V0.2 \text{ V}

C

2×10−3 V2 \times 10^{-3} \text{ V}

D

0.02 V0.02 \text{ V}

Step-by-Step Solution

Magnetic field B=5×10−5 TB = 5 \times 10^{-5} \text{ T}. Number of turns in coil N=800N = 800. Area of coil A=0.05 m2A = 0.05 \text{ m}^2. Time taken to rotate Δt=0.1 s\Delta t = 0.1 \text{ s}. Initial angle θ1=0∘\theta_1 = 0^\circ. Final angle θ2=90∘\theta_2 = 90^\circ. Change in magnetic flux Δϕ=NBAcos⁡90∘−BAcos⁡0∘=−NBA=−800×5×10−5×0.05=−2×10−3 weber\Delta \phi = NBA \cos 90^\circ - BA \cos 0^\circ = -NBA = -800 \times 5 \times 10^{-5} \times 0.05 = -2 \times 10^{-3} \text{ weber}. e=−ΔϕΔt=−−2×10−3 Wb0.1 s=0.02 Ve = -\frac{\Delta \phi}{\Delta t} = -\frac{-2 \times 10^{-3} \text{ Wb}}{0.1 \text{ s}} = 0.02 \text{ V}.

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