Question

A simple electric motor consists of a 220-turn coil, 4.4 cm in diameter, mounted between the poles of a magnet that produces a 96-mT field. Part A When a 15-A current flows in the coil, what is the coil's magnetic dipole moment?

Answer 1

Answer:

$M = 5.01\ A.m^2$          

Explanation:

It is given that,

Number of turns in the coil, N = 220

Diameter of the coil, d = 4.4 cm

Radius of the coil, r = 2.2 cm = 0.022 m

Magnetic field produced by the poles of magnet, $B=96\ mT=96\times 10^{-3}\ T$

Current flowing in the coil, I = 15 A

Let M is the coil's magnetic dipole moment. Its formula is given by :

$M=N\times I\times A$

$M=220\times 15\times \pi (0.022)^2$

$M = 5.01\ A.m^2$

So, the coil's magnetic dipole moment is $5.01\ A.m^2$. Hence, this is the required solution.