Question

A 100 turns closely wound circular coil of radius 10 cm carries a current of 3.2 A. A uniform horizontal magnetic field of strength 2T is turned on. The axis of the coil is initially in the direction of the field. The coil rotates through an angle of 90${}^{\circ }$under the influence of the magnetic field. The moment of inertia of the coil is 0.1 kg $m^2$. What is the field at the centre of the coil?

Answer 1

Magnetic field of center of coil =$N\times$ file due to one coil.

$n=$no of turns.

$=100\times \frac{{\mu }_{0}I}{2R}=100\times \frac{4\times {10}^{-7}\times 3.2}{2\times {10}^{}\times {10}^{-2}}=\frac{4\pi \times 3.2}{2\times 10}\times {10}^{-3}=2\times {10}^{-3}T=2mT$