Electro Magnetic Induction And Alternating Currents Question 477

Question: A straight conducting metal wire is bent in the given shape and the loop is closed. Dimensions are as . Now the assembly is heated at a constant rate  dT/dt=lC/s . The assembly is kept in a uniform magnetic field B=1 T, perpendicular into the paper. Find the current in the loop at the moment, when the heating starts. Resistance of the loop is 10Ω at any temperature. Coefficient of linear expansion α=106/oC .

Options:

A) 1.5×106,A anticlockwise

B) 1.5×106,A clockwise

C) 0.75×106,A anticlockwise

D) 0.75×106,A clockwise

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Answer:

Correct Answer: A

Solution:

  • Rate of change of area of the loop dAdt=A ,

    βdTdt=A.(2α)dTdt=34×2×106×1 =11.5×106m2/s

    emf=dϕdt=β.dAdt=1.5×106V current in the loop =1.5×106A The direction will be anticlockwise as the induced current will try to negate the increase in fluix due to increase in area.



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