Q. A circular coil of 20 turns and radius 10 cm is placed in uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in coil is 5 A, then the torque acting on the coil will be
A) 31.4 Nm
B) 3.14 Nm
C) 0.314 Nm
D) zero
Explanation
(d) Torque $(\tau)$ acting on a loop placed in a magnetic field $B$ is given by
$\tau = nBIA\sin\theta$
Where $A$ is area of loop, $I$ the current through it, $n$ the number of turns, and $\theta$ the angle which axis of loop makes with magnetic field $B.$
Since, magnetic field $(B)$ of coil is parallel to the field applied, hence $\theta = 0{^\circ}$ and $\sin{0{^\circ} = 0}$
$\therefore\ \ \tau = 0$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.