Question

A 200 turn coil having an axial length of 30 mm and a radius of 10 mm is pivoted in a magnetic field having a flux density of 0.8 t. If the coil carries a current of 0.5 a, the torque acting on the coil will be

Answer 1

Answer: $\tau = 0.048 N-m$

Explanation:

Given that,

Number of turns N = 200

radius r = 10 mm

length l = 30 mm

magnetic field B = 0.8 T

Current = 0.5 A

We know that,

The torque is

$\tau = 2BLINr$

$\tau = 2\times 0.8\ T\times 0.5\ A\times 30\times10^{-3}\ m\times200\times 10\times10^{-3}\ m$

$\tau = 0.048 N-m$

Hence, the torque acting on the coil will be 0.048 N-m.