Question

Chapter-21; Magnetic Fields and Magnetics Forces. 3. A circular loop of wire with N= 40 turns and of radius r = 0.45 m (cross-sectional area A = ir) carries a constant current of I = 6 A. The loop may be rotated about an axis that passes through the center and lies in the plane of the loop. When the orientation of the normal to the loop with respect to the direction of the magnetic field is o = 60°, the torque on the coil is t = 67 N. m. What is the magnitude of the uniform magnetic field B exerting this torque on the loop? A) 0.30 T B) 0.40 T C) 0.50 T D) 0.60 T E) 0.80 T

Answer 1

Answer: GYN: NL er Guixn: Numbu of turns of circular loop, N=40 radius of the circular loop, Y = 0.45 m Area of the loop, A- TIY?. 1 (0.45 ) A = 0.636 m2 Cumint, I = 6A angle, $ = 60° Tongus on the coil is t = 67 N.m as the torque experienced by a current loop in a unit magnetic field is given by I= NIBASin — Whore N = Number 87 tums, I = Current, B = magnetic field and A = area, Tu torque From (1) B = ? 1. NI Asing 8= 67 Nm 40X (CA) (0.636 m ) X Sinco B= 0.506 Tesla option (o is correct