TIME
OF
COMPLETION_______________
NAME____SOLUTION_________________________
DEPARTMENT OF NATURAL SCIENCES
PHYS 1112, Exam 2
Version 1
Total Weight: 100 points
Section 1
April 2, 2013
1.
Check your examination for completeness prior to starting. There are a total of ten (10)
problems on seven (7) pages.
2.
Authorized references include your calculator with calculator handbook, and the
Reference Data Pamphlet (provided by your instructor).
3.
You will have 75 minutes to complete the examination.
4.
The total weight of the examination is 100 points.
5.
There are six (6) multiple choice and four (4) calculation problems. Work five (5)
multiple choice and four (4) calculation problems. Show all work; partial credit will be given for
correct work shown.
6.
If you have any questions during the examination, see your instructor who will
be located in the classroom.
7.
Start:
Stop:
12:45 p.m.
2:00 p.m.
PROBLEM
POINTS
1-6
25
7
20
8
15
9
20
10
20
TOTAL
100
PERCENTAGE
CREDIT
CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN
MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK
FOR PARTIAL CREDIT.
1. An electron moves in the plane of this paper toward the top of the page. A magnetic field
is also in the plane of the page and directed toward the right. The direction of the
magnetic force on the electron is described by which of the following?
a. Toward the right edge of the page.
b.
Downward into the page.
c.
Toward the top of the page.
d.
Upward out of the page.
e.
Toward the left edge of the page.
f.
Toward the bottom of the page.
(5)
2. A charged particle is moving perpendicular to a magnetic field in a circle with a radius r.
The magnitude of the magnetic field is increased. Compared to the initial radius of the
circular path, the radius of the new path is which of the following?
a. Larger.
b. Smaller.
(5)
c.
Equal in size.
3.A charged particle is traveling through a uniform magnetic field. Which of the
following statements are true of the magnetic field? (Select all that apply.)
a. It does not change the magnitude of the momentum of the particle.
b. It exerts a force on the particle along the direction of its motion.
(5)
c. It increases the kinetic energy of the particle.
d. It exerts a force on the particle parallel to the field.
e. It exerts a force that is perpendicular to the direction of motion.
4. A loop carrying a current is placed into the uniform magnetic field. The torque that the
magnetic field exerts on the coil does NOT depend on
a. The magnitude of the current.
b. The strength of the magnetic field.
(5)
c. The number of turns in the loop.
d. The shape of the loop.
.
5. Two long, parallel wires each carry the same current I in the opposite direction (see
figure below). The magnetic force between the wires is
a. Zero.
b. Attractive.
(5)
c. Repulsive.
6. What happens to the magnitude of the magnetic field inside a long solenoid if the current
is halved?
a. It becomes four times larger.
b. It becomes twice as large.
(5)
c. It is unchanged.
d. It becomes one-half as large.
PHYS 1112 Exam 2, Version 2
Spring 2013
3
7. An electron, a proton, and a neutron each has a speed of 1.2 x 106 m/s. They are projected
into a semi-infinite region with a uniform magnetic field of magnitude 50.0 mT directed
as indicated below. (Hint: me = 9.11 x 10-31 kg, mp = 1.67 x 10-27 kg, mn = 1.67 x 10-27
kg.)
a. Schematically indicate the path that each particle follows.
Neutron goes straight through the magnetic field without deviation, electron deviates toward the
top of the page (circles counterclockwise), proton deviates toward the bottom of the page
(circles clockwise).
b. Determine the radius of the path followed by an electron.
r
mV
(9.11  1031 kg)(1.20  106 m / s )

 0.137  103 m
19
3
|q|B
(1.60  10 C )(50.0  10 T )
c. Determine the radius of the path followed by a neutron.
Straight path, r = ∞
8. A wire of infinite length carries a current of 10.0 A in a uniform magnetic field of magnitude
0.350 T as indicated below.
PHYS 1112 Exam 2, Version 2
Spring 2013
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a. Find the force on a one meter of the wire. (Specify both magnitude and
direction.)
F  IBl sin   (10.0 A)(0.350 T )(1.00 m) sin(20.0o )  1.20 N out of the page
b. If the angle that the wire makes with the magnetic field is doubled, what happens
to the direction of the force?
Nothing, still out of the page
c. Is the magnitude of the force doubled? Explain your answer.
No, not quite, since the force is proportional to the sine of the angle, not the angle itself.
9. A square, single-turn wire loop 1.00 cm on the side is placed inside a solenoid that has a
radius of 3.00 cm and is 15.0 cm long. The solenoid has 120 turns.
Here you are looking into the
solenoid.
PHYS 1112 Exam 2, Version 2
Spring 2013
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a. If the current in the solenoid is 2.00 A counterclockwise, what is the magnitude and
direction of the magnetic field inside the solenoid?
B  0nI  (4  107 T  m / A)(120 turns / 0.150 m)(2.00 A)  2.01  103 T , out of the page
b. There is a current of 4.00 A counterclockwise in the square loop. Find the torque acting
on the loop.
  NBIA sin   (1)(2.01  103 T )(4.00 A)(0.01 m)2 sin(0o )  0
10. Given the network, write equations that would allow you to solve for the currents in each
resistor. Label and indicate your choices for current directions. DO NOT SOLVE THE
EQUATIONS. You MAY nevertheless solve the equations for the extra 5 points, if correct.
Assume R1 = 4.00 Ω, R2 = 8.00 Ω, and V = 12.0 V.)
a.
PHYS 1112 Exam 2, Version 2
Spring 2013
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Different sets of equations are possible depending on the choice of currents and loops. One
example is below: (I1 is in the leftmost branch, upward. I2 is in the rightmost branch, upward, I3
is in the central branch, downward.) Left loop – counterclockwise direction, right loop –
clockwise direction.
I1  I 2  I 3
4  I 3  5I 3  4 I 1  0
W  qV  (2.00  106 C )(8.60  104V  1.09  105V )  0.0450 J
4  I 3  5I 3  3I 2  8I 2  12  0
PHYS 1112 Exam 2, Version 2
Spring 2013
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