Answer:
(a) The maximum torque acting on the rotor is 7.40 x 10⁻⁴ N.m
(b) The peak power output of the motor is 0.300W
(c) The amount of work performed by the magnetic field on the rotor in every full revolution is 2.95x10⁻³ J
(d) The average power of the motor is 0.177 W
Explanation:
(a)
[tex]T_{max} = 84 (11.1*10^{-3} A) (0.0245) (0.044) (0.800T) Sin 90^{0}[/tex]
= 7.40 x 10⁻⁴ N.m
(b)
[tex]P_{max} = T_{max}w[/tex] = (7.40x10⁻⁴)(3750)[tex]( \frac{2\pi rad}{1 rev} ) ( \frac{1 min}{60 s} )[/tex]
= 0.300 W
(c)
In one half revolution, the work is
[tex]W = U_{max} - U_{min} = uBcos180^0 - (-uBcos0^0) = 2uB[/tex]
= 2NIAB = 2(7.40x10⁻⁴) = 1.48x10⁻³ J
In one full revolution, W = 2(1.48x10⁻³)
= 2.95x10⁻³ J
(d)
Time for one revolution is Δt = 60/3600 = 1/60s
[tex]P_{avg}[/tex] = W / Δt
= 2.95x10⁻³ / (1/60)
= 0.177 W
