The 1500 turn coil in a dc motor has an area per turn of 1.1 times 10^-2 m^2. Th
ID: 1552266 • Letter: T
Question
The 1500 turn coil in a dc motor has an area per turn of 1.1 times 10^-2 m^2. The design for the motor specifies that the magnitude of the magnitude of the maximum torque is 5.8 NN when the coil is placed in a 0.24 T magnetic field. What is the current in the coil. It wire has a length of 6.60 times 10^-2 and is used to make a circular coil of one turn. There is a current of 4.80 A in the wire in the present of a 1.50 T magnetic field, what is the max torque that this coil can experience? You have a wire of length L = 1.25 m from union to make the square coil of a dc motor. The action in the coil is 1 = 29 A, and the magnetic field of the motor has a magnitude of B = 0.31 T, find the max, torque exerted on the coil with the wire is used to make a single turn square coil and a two-turn square coil (Enter the magnitudes) Single turn N middot m Two turn N middot mExplanation / Answer
8.As geomrtry of the coil is not given.Since, torque is not proportional to area alone. In other words more information on the geometry of coil is essential to calculate the current in the coil.
Let geometry of the coil to be square.
If it was square the length of each side is sqrt( A) where A is area of cross section of the coil( which is assume to be square).
And the torque of two of the sides must be zero.
The remaining two sides give a torque of T and Let I be current in the coil.
Since, force experienced by the coil is given by F = B I L * n
and torque, T = B I L *n * L/2 ( the radius of rotation is half the length of the side)
n is 1500 as there are a total of 1500 number of turns in the field.
I = T *2/ ( B * n * L^2)
= T * 2 / (B * n * A)
=5.8*2/(0.24 * 1500 * 1.1 * 10^ (-2))
= 2.93 A
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