A straight wire AB of mass 40 g and length 50 cm is suspended by flexible leads in a uniform magnetic field B = 0.40 T. Find the current required to make the tension in the leads zero. Take g = 10 m s⁻².

Question

G1 Physics · Accepted Answer

For zero tension, the upward magnetic force balances weight: IlB = mg Thus, I = mg/(lB) Given: m = 40 g = 0.040 kg l = 50 cm = 0.50 m B = 0.40 T I = (0.040 × 10)/(0.50 × 0.40) = 0.40/0.20 = 2 A Final answer: 2.