A magnetic needle lying parallel to a magnetic field requires (W ) units of work to turn it through (60^ ). The torque required to maintain the needle in this position will be

Step-by-step solution

According to the question, work done required to rotate magnetic needle from ( ₁=0^ ) to ( ₂=60^ ). (W=M B ( ₁- ₂ ) ) ( aligned & =M B ( 0^ - 60^ ) & =M B (1- 1 2 ) W & = M B 2 M B=2 W (i) Torque, & =M B 60^ =M B 3 2 & =2 W 3 2 [from Eq. (i)] & =W 3 = 3 W aligned )