The 1200turn coil in a dc motor has an area per turn of 11x1

The 1200-turn coil in a dc motor has an area per turn of 1.1x10-2 m 2 . The design for the motor specifies that the magnitude of the maximum torque is 5.8 Nm. when the coil is placed in a 0.20-T magnetic field. What is the current in the coil?

Solution

The torque on a coil is given by:

= niABsin

where i = current in the coil

A = area of the coil

B = magentic field

n = number of turns of the coil

and = angle between area vector A and the field B

maximum torque would be when = 90⁰, making sin = 1

So maximum torque = niAB

or 5.8N-m = (1200) i(1.1x10^-2m_²)(0.20-T)

or current in the coil i = 2.196A

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