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.

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

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles