A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 10³ G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g ?
(If sides of the loop =10 cm, g=10 ms⁻²)

Solution

Physics sir · 3 steps

Only the top side feels a net push, so set BIL equal to mg, find I, then use V = IR.

1. Which sides matter. The two vertical sides are partly inside the field, but they carry current in opposite directions. Their horizontal forces are equal and opposite and cancel. Only the top side, fully in the field, gives a net vertical force F = BIL.
2. Convert units. B = 10³ G = 10³ × 10⁻⁴ T = 0.1 T, L = 0.1 m, and the weight is mg = 10⁻³ × 10 = 10⁻² N.

   BIL = mg ⇒ I = mg/BL = 10⁻²/(0.1 × 0.1) = 1 A

3. So. The voltage needed to drive this current through 10 Ω is V = IR = 1 × 10 = 10 V, option (c).

The idea to keep. For a loop hanging partly in a field, forces on the side wires cancel in pairs. Only the side fully inside the field gives BIL. And always convert gauss to tesla: 1 G = 10⁻⁴ T.

Answer from the NTA final answer key. Solution by PrepWiser.