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 10 3 G , directed outwards in the shaded region. A

Question

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 $10^{3} G$ , 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 $= 10 cm, g = 10 m s^{- 2}$ )

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TestHub · Accepted Answer

For balancing of force, Magnetic force F m = weight of supporting mass m g . ∴ I L B = m g ∴ V R L B = m g ∵ From Ohm's law , I = V R ∴ V = m g R L B = 1 × 10 - 3 kg 10 m s - 2 10 Ω 0 . 1 m 10 3 × 10 - 4 T = 10 V

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