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← JEE Main 2026 · 4 Apr · Shift 2
PhysicsInteger+4 / -1Question 49 of 75

JEE Main 2026 · 4 Apr · Shift 2, Physics question 49

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is α×10−4\alpha\times10^{-4} N.m. The value of α\alpha is ________. (π=3.14\pi=3.14)

Answer and worked solution

Official answer: 314

Torque.
τ=NIABsin⁡30∘=125(1)(3.14×4×10−4)(0.4)(0.5)=0.0314 N m=314×10−4 N m.\tau=NIAB\sin30^\circ=125(1)(3.14\times4\times10^{-4})(0.4)(0.5)=0.0314\text{ N m}=314\times10^{-4}\text{ N m}.

Answer: 314