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