Download AP Physics – Magnetic Electric - 5

AP Physics – Magnetic Electric - 5
1. Why would banging on a magnet with a hammer cause it to lose its magnetism? Banging on it
rearranges the individual atoms and their magnetic fields. Banging on it also causes heat. Heat allows
the atoms in a magnet to become less organized, causing the individual magnetic fields of the atoms to
cancel each other out, dissipating the effect.
FYI-An electromagnet is simply a wire coil energized by an electric charge, which align its atoms
(polarize them) as electrons flow through it, creating a temporary magnetic field.
2. A 34.5 cm wire carrying a current of 12.0 A passes through a magnetic field of 2.35 x 102 T. (a) What is
the maximum force exerted on the wire? (b) If the wire makes an angle of 35.0 to the field, what is the
force exerted on it?
FMax  BIl  2.35 x 102 T 12.0 A  0.345 m  
973 N
F  BIl sin 
 2.35 x 102 T 12.0 A  0.347 m  sin 35.00

558 N
3. A conductor has 2.35 x 1020 electrons passing through a 76.0  resistor in 1.13 s. Find (a) the current
through the resistor and (b) what is the potential difference across the resistor?
q
I
t
2.35 x 1020 e   1.6 x 1019 C 


 

1.13 s
1
e


V  IR  33.3 A  76.0   
3.33 x 101 A or 33.3 A
2530 V
4. A Certain light bulb is designed to dissipate 6.00 W when it is connected to a 12.0 V source.
(a) Calculate the resistance of the light bulb.
P
V
6.0 W
12.0 V
P  IV I 

 0.50 A
V  IR R 


24.0 Ω
V
I
12.0 V
0.50 A
(b) If the light bulb functions as designed and is lit continuously for 30 days, how much energy is used?
 24 h   3600 s 
W
J
P
W  Pt  6.00  30 day  
1.56 x 107 J

  15600000 J 
t
s
1
day
1
h



The 6.00 W, 12.0 V bulb is connected in a circuit with a 1 500 Watt, 120 V toaster; an adjustable resistor;
and a 120 V power supply. The circuit is designed so that the bulb and the toaster operate at the given
values and, if the light bulb fails, the toaster will continue to function at these values.
(c) On the drawing below draw in wires connecting the components shown to make a complete circuit
that will function as described above.
120 V
Power supply
12 V bulb
Adjustable Resistor
Toaster
(d) Determine the value of the adjustable resistor that must be used in order for the circuit to work as
designed.
V  Vbulb
V  Vbulb  VRe sistor
V  Vbulb  IR
R
I
P
6.00 W
120 V  12.0 V
216 
P  IV
I

 0.500 A

R
12.0 V
0.500 A
V
(e) If the resistance of the adjustable resistor is increased, what will happen to the following?
i.
The brightness of the bulb – explain your response.
ii.
Dims, less current, less power
The power dissipated by the toaster. Explain your reasoning.
Nothing it’s in parallel and still sees 120 V.
5. A Circuit is set up as shown. (a) What is the current? (b) What is the total resistance? (c) What is the
power dissipated by the 10.0  resistor?
1
1
1
1
1
18.0 W
65.0 W
(b)




R234  12.5 
15  65   10 
R 234 R2 R3  R4
15.0 W
15.0 V
R  18   12.5   12  
(a) I 
V
R

15.0 V
42.5 
(c) V234  IR234
I 34 
V
R34
P  IV

12.0 W

0.353 A
 0.353 A 12.5   
4.41 V
65   10 
10.0 W
42.5 
4.41 V
V4  I 34 R4
 0.0588 A
  0.0588 A  0.588 V

 0.0588 A 10.0    0.588 V
0.0346 W
6. An electron is accelerated through 2400 V from rest and then enters a region in which there is a uniform
1.70 T magnetic field. What are (a) the maximum and (b) the minimum magnitudes for the magnetic
force this charge can experience?
1
E  qV  mv 2
2
2qV
v
m
9.11 x 1031 kg
m2
m
 2.90 x 107
2
s
s
m

FMax  qvB  1.6 x 1019 C   2.90 x 107  1.70 T  
7.89 x 1012 N
s

F min is zero
v   0.843 x 1015
m2
s2

2 1.6 x 1019 C  2.4 x 103V
 8.43 x 1014
max force
7. A duck flying horizontally due north at 15 m/s passes over Atlanta, where the magnetic field of the earth
is 5.0 x 10-5 T in a direction 60.0 below a horizontal line running north and south. The duck has a
positive charge of 4.0 x 10-8 C. What is the magnetic force acting on the duck?
 m
F  qvB sin   4.0 x 108 C 15   5.0 x 105 T  sin 600  260 x 1013 N 
2.6 x 1011 N
 s