The figure shows a square loop L of side 5   c m which is connected to a network of resistances. The…

The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards the right with a constant speed of 1 cm s-1 . At some instant, a part of L is in a uniform magnetic field of 1T perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω, the current in the loop at that instant will be close to:
  1. 115 μA
  2. 60 μA
  3. 150 μA
  4. 170 μA

Solution

If B is magnetic field, v is velocity of rod and l is the length of rod, then induced EMF in moving road is given by 

    ε=Bvl     
    =1T(1cm s-1)(5cm)
    =1×1×10-2×5×10-2
    =5×10-4  V
As given circuit forms balanced whetstone bridge, the current through 3Ω is zero.
Equivalent resistance Req=4×24+2+1.7=3 Ω
Current in the circuit i=εReq=5×10-43
=167 μA170 μA

Asked in: JEE Main 2019 (12 Apr Shift 1)

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