Physics
HS/Science
Unit: 10 Lesson: 01
KEY
Answer the questions on separate paper or in your science notebooks.
Equations that may prove useful:
Emf
V – Emf
V rms max
= N B A ω backc
= V max
= I run
R
Maximum waveform voltage for generator
Equation relating input voltage, back emf and current for motor
(.707)Relationship between maximum waveform voltage and rms value
V
2
= (N
2
P = VI = I
Φ = BA
/N
2
1
)V
R
1
Transformer equation solved for output voltage
Familiar power equations
Definition of magnetic flux
Emf = N ΔΦ/Δt
Emf = Blv
Faraday’s law
Voltage induced in wire moving through a magnetic field.
1.
A 240 V dc motor has an armature (rotating coil) whose resistance is 1.50 Ω. When running at its operating speed, it draws a current of 16.0 A.
(a) What would be the starting current (current when the coil is not rotating)?
(b) When running, the motor induces (generator effect) a back emf to limit the current to 16 A.
What is the back emf of the motor when it’s running? (Hint: 240-back emf = IR when running)
V = 240 V
I
R = 1.50 Ω run
= 16.0 A
I start
= V/R = 240/1.5 = 160A a. I start
= 160 A V – Emf backc
= I run
R: Emf back
= V – I run
R: b. Emf back
= 216 volts Emf back
= 240 – 1.5(16) = 216
2. Consider the two coils to the right. A current flows through the outer coil in a counter clockwise fashion but decreasing. In what direction is the induced current in the inner coil?
Counter clockwise: It tends to keep the magnetic field from decreasing inside the loop.
3. The coil of a generator has 50 loops of cross sectional area of 0.25
m 2 . What is the maximum emf generated by this generator if it is spinning with an angular velocity of 4.0 rad/s in a 2.0 T magnetic field?
N = 50
A = 0.25 m 2
Emf max
= 100 volts
ω = 4.0 rad/sec
B = 2.0 T
Emf max
= NBAω = 50 (2)(.25)(4) = 100 volts
4. What would be the rms value of the voltage for the generator in problem 3? What value would a voltmeter measure?
Emf max
= 100 volts V rms
= V max
(.707) = 70.7 volts
V rms
= 70.7 volts
©2012, TESCCC 11/07/12 page 1 of 3

Physics
HS/Science
Unit: 10 Lesson: 01
5. An ideal transformer has 60 turns on its primary coil and 300 turns on its secondary coil. If 120-V at 2.0 A is applied to the primary, what voltage and current are present in the secondary? What power is supplied to the transformer?
N
1
= 60
N
2
= 300 Power = VI = (120)(2) = 240 watts
V
1
I
1
= 120 V
= 2 A
V
2
= 600 volts
I
2
= 0.4 amps
Power = 240 watts
V
2
= (N
I
2
= (N
1
2
/N
1
)V
/N
2
) I
1
1
= (300/60) 120 = 600 volts
= (60/300) 2 = 0.4 amps
6. A boy drops a magnet down a copper tube, and a girl drops an identical strong magnet down a plastic tube at exactly the same time. Which magnet hits the floor first and why?
The magnet in the plastic tube hits first. The copper tube has currents induced (ghost magnets) to slow down the magnet in the copper tube.
7. A clever but not well-schooled inventor decides that a magnet on a spring could be used to power a light bulb. Current is generated as the magnet oscillates back and forth through the coil. Why will this not work? Or will it?
This would work to power the light if the hand kept supplying energy.
If simply allowed to oscillate without additional energy input, the magnet oscillating magnet would quickly come to rest. The induced current in the spring (ghost magnet) would (like in the copper tube above) slow the motion of the oscillating magnet.
8. The voltage of most wall plugs in the home or at school is rated at 115-V – 60 Hz. What is the amplitude voltage of these signals? What would be the amplitude value of a current of 1 Amp?
The quoted voltage and current values are rms values. Thus, the amplitude values are larger by a factor of 1.41.
115 volts has an amplitude voltage of 162 volts.
The amplitude current of 1 amp is 1.41 amps.
9. An ideal transformer has 160 turns on its primary coil and 320 turns on its secondary coil. If 120-V at 2.0 A is applied to the primary, what voltage and current are present in the secondary?
N
N
2
1
= 160
= 320
V
1
I
1
= 120 V
= 2 A
V
2
= 240 volts
I
2
= 0.5 amps
V
2
= (N
I
2
= (N
1
2
/N
1
)V
/N
2
) I
1
1
= (320/160)120 = 240 volts
= (160/320)2 = 0.5 amps
10. The magnetic flux through a coil changes from 4 x 10 -5 Wb to 5 x 10 -5 Wb in 0.1 seconds. What emf is induced in this coil?
Ф
Ф
1
2
= 4 x 10
= 5 x 10
-5
-5
Δt = 0.1 sec
Wb
Wb emf = (Ф
2
- Ф
1
)/Δt = (5 x 10 -5 – 4 x 10 -5 )/.1 = 1 x 10 -4 volt
©2012, TESCCC 11/07/12 page 2 of 3

Physics
HS/Science
Unit: 10 Lesson: 01 emf = 1 x 10
11. A 0.4 m 2 coil, consisting of 100 turns, flips completely 360 degrees in a magnetic field of 0.05 T in a time of 0.1 seconds. What is the average emf induced in the coil? What is the peak emf if this flip has a constant rotational speed during the flip? (Hint: Consider a generator cycle for the peak emf.)
The average induced emf for the full 0.1 seconds is zero, since the flux does not change.
However, the peak emf can be found from the generator emf equation.
A = 0.4 m 2
N = 100
ω = 2π radians/0.1sec = 62.8 rad/sec
B = 0.5 T
Emf max
= NBAω = 100(.5)(.4)(62.8) = 1256 volts
Emf max
= 1256 volts
12. The secondary coil of a neon sign transformer provides 7500 volts at 0.10 A. The primary coil operates on 120 volts. What current does the primary draw?
I
V
2
= 7500 volts
2
V
I
1
= 0.10 amps
1
= 120 V
= 6.25 A
The Power In = Power Out = VI = (7500)(.1) = 750 watts
Thus,
I
1
= P/V
1
= 750/120 = 6.25 amps
13. Describe how a dc generator operates, and include how the ac current is converted to dc and the resulting waveform output of the dc voltage.
A typical dc generator works by rotating a coil between the poles of a permanent magnet.
As the coil turns, an ac signal is generated each half turn. At the end of the half turn, a split ring commutator reverses the direction of the ac voltage produced, and instead of the sine wave of an ac generator, a humped output is delivered.
©2012, TESCCC 11/07/12 page 3 of 3