Question

1. Consider a generator that rotates its 200 turn, 0.13 m diameter coil at 3550 rpm in a 0.65 T field.

a. Calculate the peak voltage of the generator.

2. A precision laboratory resistor is made of a coil of wire. The coil is 1.5 cm in diameter, 3.75 cm long, and has 500 turns.

a.

What is its self-inductance in mH?

b. What average emf in V is induced if the 12.0 A current through it is turned on in 5.00 ms (one-fourth of a cycle for 50 Hz AC)?  

Answer 1

1)Peak voltage=BAN$\omega$

B=magnetic field=.65T

A=area of cross section of coil=
77
=3.14X
10652
=.01327m2(r=radius=.13/2=.065m

N=number of turns=200

$\omega$=angular velocity=2$\pi$f

f=frequency of rotation=3550rpm=3550/60=59.17Hz

$\omega$=2X3.14X59.17=371.59rad/s

Peak voltage=BAN$\omega$=.65X.01327X200X371.59=641.03volt

2

a)Self inductance L=
HO

N4
A/l

HO
=4$\pi$X
10:

N=number of turns=500

A=area of cross section=$\pi$$r^{2}$=3.14X$.0075^{2}$=1.77X$10^{-4}$m2 (r=1.5/2=.75cm)

l=length=3.75cm=.0375m

L=
HO

N4
A/l=4$\pi$X
10:
x$500^{2}$X1.77X$10^{-4}$/.0375=1.48mH

b)induced emf e=Ldi/dt

di=change in current=12-0=12A

dt=time=5ms=.005

L=1.48mH=.00148H

e=Ldi/dt=.00148x12/.005=3.552volt