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Question

Physics Question on Electromagnetic induction

A circular disc of radius 0.2m0.2\, m is placed in a uniform magnetic field of induction (1π)Wbm2(\frac{1}{\pi})Wb\,m^{-2} in such a way that its axis makes an angle of 6060^{\circ} with B\vec{B}. The magnetic flux linked with the disc is

A

0.02Wb0.02\, Wb

B

0.06Wb0.06\, Wb

C

0.08Wb0.08\, Wb

D

0.01Wb0.01\, Wb

Answer

0.02Wb0.02\, Wb

Explanation

Solution

Here, r=0.2mr =0.2\,m, B=1πWbm2B =\frac{1}{\pi}\,Wb\,m^{-2}, θ=60\theta=60^{\circ} \therefore Magnetic flux, ϕ=BAcosθ=B(πr2)cosθ\phi=BA\, cos\,\theta=B(\pi\,r^{2})cos\, \theta =1ππ(0.2)2cos60=\frac{1}{\pi} \pi(0.2)^{2} \, cos\, 60^{\circ} =0.02Wb=0.02\, Wb