An electric motor has a square coil measuring 5.0 cm on each side and consisting

Question

An electric motor has a square coil measuring 5.0 cm on each side and consisting of 15 windings. The permanent magnets in the motor produce a uniform magnetic field of 1.2 T. If the current through the coil is 0.50 A, (a) how much torque is generated when the axis of the coil, (otherwise known as the normal of the coil, perpendicular to the plane of the coil), points 35 away from the direction of the magnetic field? Include a diagram. (b) What are the minimum and maximum instantaneous torques generated by the motor?

Explanation / Answer

a)

Here ,

Magnetic moment = area*N*I

Magnetic moment = 0.05^2 * 15*0.50

Magnetic moment =0.01875 A.m^2

Therefore , Torque = B*u*sin(theta)

Torque = 0.01875 * 1.2 * sin(35)

Torque = 0.01291 N.m

the torque is 0.01291 N.m

b)

Here ,

Maximum torque is

Tmax = 0.01875 * 1.2

Tmax = 0.0225 N.m

Maximum torque is 0.0225 N.m

and mimimum torque is 0 N.m

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