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 1900 rpm. What do you specify for the alternator's magnetic field?

Answer 1

The alternator's magnetic field is 0.036 T

Step-by-step explanation:

The area of the rotating coil is equal:

`A=(\pi d^(2) )/(4)`

Where

d = 10 cm = 0.1 m

`A=(\pi 0.1^(2) )/(4) =0.0079m^(2)`

The angular frequency is:

w = 1900 rpm = 198.97 rad/s

The emf is equal:

`B=(E)/(NAw)`

Where

E = 14 V

N = 250 turns

`B=(14)/(250*0.0079*198.97) =0.036T`