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Strike441 [17]
3 years ago
10

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 9

6-mT field. Part A When a 15-A current flows in the coil, what is the coil's magnetic dipole moment?
Physics
1 answer:
Verdich [7]3 years ago
4 0

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.

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