The rotor in a certain electric motor is a flat, rectangular coil with 76 turns

Question

The rotor in a certain electric motor is a flat, rectangular coil with 76 turns of wire and dimensions 2.50 cm by 4.00 cm. The rotor rotates in a uniform magnetic field of 0.800 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, it carries a current of 5.3 mA. In this orientation, the magnetic moment of the rotor is directed opposite the magnetic field. The rotor then turns through one-half revolution. This process is repeated to cause the rotor to turn steadily at 3600 rev/min.

Explanation / Answer

t = NIAB * sin theta

for maximum theta = 90 degree

tmax = N*I*A*B

N = 76 turns

A = (0.025 x 0.04)

tmax = 3.2 x 10^-4 N-m

part b )

Pmax = tmax * w

w = 3600 rev/min = 3600 * 2pi rad/ 60 s

Pmax = 0.1206 W

part c ) in one half revolution

W = Umax - Umin

U = -mu*Bcostheta

Umax = -mu*Bcos180

Umin = -mu*Bcos0

W = 2muB

mu = magnetic dipole moment = NIA

W = 2*N*I*A*B

W = 2 * 76 * 5.3 x 10^-3 * 0.800

W = 6.44 x 10^-4

in full W = 2W = 1.28 x 10^-3 J

part d )

Pavg = W/t

t = 1/60s

Pavg = 1.28 x 10^-3 / (1/60)

Pavg = 0.07728 W

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