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Question

Physics Question on Electromagnetic induction

A 800 turn coil of effective area 0.05 m2m^2 is kept perpendicular to a magnetic field 5×1055\times 10^{-5} T. When the plane of the coil is rotated by 9090^\circ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be

A

2×1032\times 10^3V

B

0.02 V

C

2V

D

0.2 V

Answer

0.02 V

Explanation

Solution

Magnetic field B=5×105  TB = 5 \times 10^{-5} \; T
Number of turns in coil N = 800
Area of coil A = 0.05 m2m^2
Time taken to rotate Δ\Deltat = 0.1 s
Initial angle θ1=0\theta_1 = 0^{\circ}
Final angle θ2=90\theta_2 = 90^{\circ}
Change in magnetic flux Δϕ\Delta \phi
=NBAcos90BAcos0= NBA \cos 90^{\circ} - BA \cos 0^{\circ}
= - NBA
=800×5×105×0.05= - 800 \times 5 \times 10^{-5} \times 0.05
= - 2 ×103\times 10^{-3} weder
e=ΔϕΔt=()2×103Wb0.1s=0.02Ve =- \frac{\Delta\phi}{\Delta t} = \frac{-\left(-\right)2\times10^{-3}Wb}{0.1 s} = 0.02 V