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Question:

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 right with a constant speed of 1 cm s⁻¹. At some instant, a part of L is in a uniform magnetic field of 1 T, 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:

60 μA

115 μA

170 μA

150 μA

Solution:

Correct option is B. 170 μA
Since it is a balanced wheatstone bridge, its equivalent resistance = 4/3 Ω
ε = Blv = 5 × 10⁻⁴ V
So total resistance R = 4/3 + 1.7 ≈ 3 Ω
∴ i = ε/R ≈ 166 μA ≈ 170 μA