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Asked in GSEB Board March 2023 · Torque and potential energy
Given: B=0.3 T along +Z; a rectangular loop 10 cm×5 cm carrying I=12 A, lying in the XY plane as the figure shows.
Idea: the torque on a current loop is τ⃗=⃗m×⃗B, where ⃗m=I⃗A points along the normal to the plane of the loop.
Area: A=0.10×0.05=5×10⁻³ m², so m=12×5×10⁻³=0.06 A m².
The loop lies in the XY plane, so its normal is along Z: ⃗m is along the field line, here pointing the opposite way, and the angle between ⃗m and ⃗B is 180°.
τ=mB sin 180°=0.06×0.3×0, which is zero.
The loop feels no torque: it is already in equilibrium. Only a loop whose plane contained the field would feel the full mB=1.8×10⁻² N m.
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