Electromagnetic Induction Study Guide

Shared by:
Categories
Tags
-
Stats
views:
43
posted:
6/16/2012
language:
English
pages:
3
Document Sample

```							Electromagnetic Induction Review Sheet Answers                                            AP Physics

Review Questions
1. A door is rigged with a circuit that
rings a bell when the door is opened.
Pushing the door open shoves a magnet
attached to the door toward a solenoid of
20 coils with a cross-sectional area of
.0004 m2 as shown. The magnetic field
inside the coils changes from 0.05 T to
0.18 T in 1.9 s. (Assume that the magnetic
field is perpendicular to the coils and has
the same strength at all coils)
a) How big is the electromotive force between the two ends of the
solenoid generated by electromagnetic induction?
         5.2 x 10-5 Wb
E = -N     = - 20                = 5.48 x 10-4 V
t             1.9 s
1 = B1Asin = 0.05 T x .0004 m2 x sin90° = 2 x 10-5 Wb
2 = B2Asin = 0.18 T x .0004 m2 x sin90° = 7.2 x 10-5 Wb
 = 5.2 x 10-5 Wb

b) If the resistance in the bell circuit that is connected to the solenoid is 5 , how large and in what
direction is the current flowing through it?
V 5.48 x 10-4 V
I=R=                   = 1.09 x 10-4 A, going counter clockwise as seen from magnet (see picture)
5
2. A magnet is pulled away from a 2 cm
radius, 7 coil solenoid at a rate of 4 cm/s.
While the magnet moves 7.5 cm, the
magnetic field drops in strength from 0.8 T
to 0.22 T. (Assume that the magnetic field is
perpendicular to the coils and has the same
strength at all coils)
a) Which way will the current through the solenoid flow?
It will be going counter clockwise when seen from the
magnet’s end so that it the induced current tries to
support the field to the left as it weakens.

b) How big is the potential difference generated by electromagnetic induction between the two ends of
the solenoid?
 7.308 x 10-4 Wb
E=      =                  = 2.74 x 10-5 V
t       1.875 s
d 7.5 cm
t = v = 4 cm/s = 1.875 s

A =r2 =  (0.02 m)2 = 0.00126 m2
1 = B1Asin = 0.8 T x .0.00126 m2 x sin90° = 0.001008 Wb
2 = B2Asin = 0.22 T x 0.00126 m2 x sin90° = 2.772 x 10-4 Wb
 = 7.308 x 10-4 Wb
3. A portable generator runs by spinning 300 square loops of wire that are 10 cm x 10 cm through a
0.106 T magnetic field at a frequency of 60 Hz.
a) What is the maximum emf produced?
E0 = NBA = 300 x 0.01 m2 x 0.106 T x 377 rad/s = 120 V
 = 260 Hz = 377 rad/s

b) What is the average emf produced by the generator
E0 120 V
Eavg =      =      = 84.85 V
2     2

4. You have a generator that is made from 450 circular wire loops that are 3 cm in diameter and rotate
in a magnetic field of 1.7 T from permanent magnets. How quickly would the loop have to rotate to
produce an average emf of 15 V?
E0
Eavg =
2
E0 = 2 · Eavg = 2 · 15 V = 21.21 V
E0               21.21 V
 = NBA = 450 x 1.7 T x 0.00141 m2 = 19.66 rad/s
f=     =             = 3.13 Hz
2        2

A =r2 =  (0.015 m)2 = 0.00141 m2

5. When it is turning at operating speed, a motor with 7  of resistance connected to 120 V draws 2 A
of current. What is the back emf of the motor at operating speed?
Vactual = IR = 2A x 7  = 14 V
Vactual = Vinput – Eback
Eback = Vinput – Vactual = 120 V – 14 V = 106 V

6. A Jacob’s Ladder (the ubiquitous mad scientist device where an electric spark climbs between two
wires while making a buzzing sound) is based around a transformer.
a) If the transformer has 60 loops in its primary coil and 7200 loops in its secondary coil, what will the
output voltage be when you plug it into a 120 V outlet?
The outlet is the primary (input) side and the Jacob’s ladder is the secondary (output) side.
VS NS
=
VP NP
NS     7200
VS = VP N = 120 60 = 14400 V
P

b) If the primary coil has a maximum current of 10 A, what will the maximum current in the secondary
coil be?
ISVS = IPVP
VP          120 V
IS = IPV = 10 A x 14400 V = 0.083 A
S
7. The electrical lines running through the power distribution station are at 230,000 V.
a) If they wanted to tap those lines to run a coffee pot in the operating room that required 4.7 A of
current at 120 V and had a transformer with 200 turns in the secondary coil, how many turns would the
primary coil need to have?
NS VS             NP VP
NP = VP           NS = VS
VP     230000 V
NP = NS V = 200 120 V = 383000 turns
S

b) How much current would be drawn from the high voltage line?
IPVP = ISVS
VS            120 V
IP = ISV = 4.7 A x 230000 V = 0.00254 A
P

```
Related docs
Other docs by pn641q
Cap 6 Final Review