Question

An equilateral triangular coil of 10 turns carrying 1A current is placed in uniform magnetic field of 4T as shown in figure. If side of triangle is 1m then torque acting on coil is

Answer 1

$\tau =\mu Bsin\theta =niABsin\theta \tau =10\times 1\times 4\times \frac{\sqrt{3}}{4}\times 1^{2}sin{90}^{\circ }$
$\tau =1.73\times 10=17.3$