Question

You’re an electrical engineer designing an alternator (the generator that charges a car’s battery). Mechanical engineers specify a 10-cm-diameter rotating coil, and you determine that you can fit 250 turns in this coil. To charge a 12-V battery, you need a peak output of 14 V when the alternator is rotating at 1200 rpm. What do you specify for the alternator’s magnetic field? Wolfson, Richard. Essential University Physics, Volume 2 (p. 542). Pearson Education. Kindle Edition.

Answer 1

Answer:

I will specify a value of 0.009T for the alternator’s magnetic field

Explanation:

E_peak = 14 V

d = 10cm = 0.1m, so r = 0.1/2 =0.05m

N = 250 turns

f = 1200rpm = (1200rp/m x 1m/60sec) = 20 revolutions per second

At peak performance, peak voltage is given by the equation;

E_peak = NABω

Let's make the magnetic field B the subject;

B = E_peak/(NAω)

Now we know that ω = 2πf

Thus, ω = 2π x 20 revs/s = 125.664 revs/s.

Let's convert it to the standard unit which is rad/s.

1 rev/s = 6.283 rad/s

Thus, 125.664 revs/s = 125.664 x 6.283 = 789.55 rad/s

Area (A) = πr² = π x 0.05² = 0.007854 m²

Thus, plugging in the relevant values to get;

B = 14/[(250 x 0.007854 x 789.55)] = 0.009T