Question

You're an electrical engineer designing an alternator (the generator that charges a car's battery). Mechanical engineers specify a 10-cmcm-diameter rotating coil, and you determine that you can fit 250 turns in this coil. To charge a 12-VV battery, you need a peak output of 17 VV when the alternator is rotating at 1500 rpm.

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

Answer 1

13.78 mT

Step-by-step explanation:

The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,

A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength

So,

B = ε/ωNA

substituting the values of the variables into the equation given that ε = 17 V

So, B = ε/ωNA

B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)

B = 17 V/(1233.8634 rad-turns-m²/s)

B = 0.01378 T

B = 13.78 mT