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Question:

Consider a loop that is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

µ₀I²a²d

µ₀I²a²/2d

√3µ₀I²a²d

√3µ₀I²a²/2d

Solution:

T=MBsinθ=Iπa²×µ₀Iπdsin30°=µ₀I²a²/2d