Question

A simple motor has N=250 turns of wire on a coil measuring 12 cm \times \, 4 cm. It's resistance is R=10 ohms. The magnetic field is B=1 T.

(a) How much current (in Ampere) in the coil is needed to produce a maximum torque of 40 Nm?

b) What maximum EMF (in Volts) is needed to drive the motor at 50 Hz if the current is constant?

Answer 1

(a)
Torque is
M=p⒨•B•sinα
p⒨=NIA
If M(max) =sin α=1
M(max)=NIAB,
I= M(max)/NAB =
=40/250•0.12•0.04•1=33.3 A
(b)
ℰ=-dΦ=-d(NBAsinωt)/dt=
=-NBAωcosωt
ℰ(max)=NBAω=2πNBAf=
=2π•250•1•0.12•0.04•50=377 V

Answer 2

Thanks