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) 5/3V

B) 2π/3V

C) (53+2π/3),V

D) (53+π),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,

    e1=Bv(BF)=5×1×2,sin,60o=53,V

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

    e2=AdBdt=πR23(dBdt)=π(1)23×2=2π3V So, e=e1+e2=(53+2π3)V



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