Electro Magnetic Induction And Alternating Currents Question 483

Question: A cylindrical region of radius 1 m has instantaneous homogenous magnetic field of 5T and it is increasing at a rate of 2T/s. A regular hexagonal loop ABCDEFA of side 1 m is being drawn in to the region with a constant speed of 1 m/s as . What is the magnitude of emf developed in the loop just after the shown instant when the corner A of the hexagon is coinciding with the centre of the circle?

Options:

A) $ 5/\sqrt{3}V $

B) $ 2\pi /\sqrt{3}V $

C) $ (5\sqrt{3}+2\pi /3),V $

D) $ (5\sqrt{3}+\pi ),V $

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Answer:

Correct Answer: C

Solution:

  • The induced emf across the ends B and F due to motion of the loop,

    $ e_{1}=Bv(BF)=5\times 1\times 2,\sin ,60^{o}=5\sqrt{3},V $

    The induced emf across the loop due to change in magnetic field

    $ e_{2}=A\frac{dB}{dt}=\frac{\pi R^{2}}{3}( \frac{dB}{dt} )=\frac{\pi {{(1)}^{2}}}{3}\times 2=\frac{2\pi }{3}V $ So, $ e=e_{1}+e_{2}=( 5\sqrt{3}+\frac{2\pi }{3} )V $



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