A semicircular wire of radius a carries current I in the horizontal x–y plane and is hinged about the y-axis. A uniform magnetic field B⃗ = −Bî is applied. If the loop remains horizontal in equilibrium, its mass is m = π²aBI/(ng). Find n. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A semicircular wire of radius a carries current I in the horizontal x–y plane and is hinged about the y-axis. A uniform magnetic field B⃗ = −Bî is applied. If the loop remains horizontal in equilibrium, its mass is m = π²aBI/(ng). Find n. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Centre of mass of a uniform semicircular arc is at distance 2a/π from centre toward the curved arc. Thus gravitational torque about the diameter/hinge is τg = mg(2a/π). Step 2 — Force on a wire element is dF⃗ = I(dℓ⃗ × B⃗). Its moment about the hinge integrates to magnetic torque τB = I B A, with semicircular area A = πa²/2. Step 3 — At equilibrium, τB = τg: IB(πa²/2) = mg(2a/π). Step 4 — Rearrange m = π²aBI/(4g). The source writes m = π²aBI/(ng), hence n = 4.