Answer:
[tex]M = 5.01\ A.m^2[/tex]
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, [tex]B=96\ mT=96\times 10^{-3}\ T[/tex]
Current flowing in the coil, I = 15 A
Let M is the coil's magnetic dipole moment. Its formula is given by :
[tex]M=N\times I\times A[/tex]
[tex]M=220\times 15\times \pi (0.022)^2[/tex]
[tex]M = 5.01\ A.m^2[/tex]
So, the coil's magnetic dipole moment is [tex]5.01\ A.m^2[/tex]. Hence, this is the required solution.
The coil's magnetic dipole moment in the electric motor is 5.02 Am².
The given parameters;
The magnetic dipole moment is calculated as follows;
[tex]\tau = N\times I \times A[/tex]
where;
A is the area of the coil,
[tex]\tau = (220) \times (15) \times (\pi \times 0.022^2)\\\\\tau = 5.02 \ Am^2[/tex]
Thus, the coil's magnetic dipole moment in the electric motor is 5.02 Am².
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