Question

Imagine that we construct a motor using a coil consisting of 300 turns of wire wound in the form of a square 2.5 cm on a side. The permanent magnets in the motor create a field with strength 0.13 T in which the coil rotates. We would like the motor to be able to produce 5.0 W of mechanical energy when it is turning at 1200 rpm. How much current do we need to supply to the motor?

ANSWER IS 2.56 A. PLEASE EXPLAIN THOROUGHLY!!!

Answer 1

Answer 2

calculation:

1. Calculate the area of the coil:

   $$A=(2.5\times 10^{-2}{)}^2=6.25\times 10^{-4}{\text{m}}^2$$

2. Convert rpm to rad/s:

   $$\omega =1200\times \frac{2\pi }{60}=125.66\text{rad/s}$$

3. Use the power equation with average back emf:

   $$P=\left(\frac{2}{\pi }NAB\omega \right)I$$$$I=\frac{P\pi }{2NAB\omega }$$

4. Plug in the numbers:

   $$I=\frac{5.0\times \pi }{2\times 300\times 6.25\times 10^{-4}\times 0.13\times 125.66}\approx 2.56\text{A}$$

Final Answer: The motor needs to be supplied with approximately 2.56 A of current to produce 5.0 W of mechanical power at 1200 rpm

 

Why 2.56 A ?

The average torque over a full rotation is reduced by the factor $\frac{2}{\pi }$, leading to a higher required current compared to the peak torque scenario. This adjustment ensures the motor delivers 5.0 W consistently at 1200 rpm.