Answer:
The alternator's magnetic field is 0.036 T
Explanation:
The area of the rotating coil is equal:
[tex]A=\frac{\pi d^{2} }{4}[/tex]
Where
d = 10 cm = 0.1 m
[tex]A=\frac{\pi 0.1^{2} }{4} =0.0079m^{2}[/tex]
The angular frequency is:
w = 1900 rpm = 198.97 rad/s
The emf is equal:
[tex]B=\frac{E}{NAw}[/tex]
Where
E = 14 V
N = 250 turns
[tex]B=\frac{14}{250*0.0079*198.97} =0.036T[/tex]
