TIME OF COMPLETION - Clayton State University... (1)(0.300 t)(0.0600 m)2 ... -7 tm/a)(5.00 a)/ ((2 )(0.02 m)) = 0.500 x 10-4 t, ... time of completion_ name_ department of

TIME OF COMPLETION_______________

NAME_____SOLUTION________________________

DEPARTMENT OF NATURAL SCIENCES

PHYS 1112, Exam 2 Section 1

Version 1 November 1, 2004

Total Weight: 100 points

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 all 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: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

20

10

15

TOTAL

100

PERCENTAGE

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN

MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK

FOR PARTIAL CREDIT.

1. What is the rms current value for an ac current with the maximum value of 10.0 A?

a. 28.2 A.

b. 3.10 A.

(6)

c. .

d. .

2. If you look directly down on the north pole of a bar magnet, the magnetic field points

a. To the right.

b. To the left.

(6)

c. Away from you.

d. Toward you.

3. A bar magnet is falling through a loop with constant velocity. The north pole enters first.

As the south pole leaves the loop of wire, the induced current (as viewed from above)

will be:

a. Clockwise.

b. Counterclockwise.

(6)

c. Zero.

d. Along the length of the magnet.

4. How is the energy stored in a current carrying inductor related to its self-inductance, L?

a. Directly proportional to L2.

b. Directly proportional to L1/2

.

(6)

c. Directly proportional to L.

d. Inversely proportional to L.

5. The path of a charged particle moving parallel to a uniform magnetic field will be:

a. Straight line.

b. Circle.

(6)

c. Ellipse.

d. Parabola.

6. A current in a solenoid coil creates a magnetic field inside that coil. The field strength is

directly proportional to:

a. The coil area.

b. The current.

(6)

c. Both a and b are valid choices.

d. None of the above choices are valid.

7. The rectangular coil pictured below has 80 turns, is 25.0 cm wide and 30.0 cm long, and

is located in a magnetic field B = 1.40 T directed out of the page as shown, with only half

of the coil in the region of the magnetic field. The resistance of the coil is 24.0 . Find

the magnitude and direction of the current if the coil is moved with a speed of 2.00 m/s

a. to the right,

= 0, since = 0

I = 0

b. up,

= N B V L = (80)(1.40 T)(2.00 m/s)(0.250 m) = 56.0 V

I = /R = (56.0 V) / (24.0 ) = 2.33 A, clockwise

c. to the left,

= 0, since = 0

I = 0

d. down.

= N B V L = (80)(1.40 T)(2.00 m/s)(0.250 m) = 56.0 V

I = /R = (56.0 V) / (24.0 ) = 2.33 A, counterclockwise

8. A current-carrying wire is bent into a shape of a square f sides L = 6.00 cm and is placed

in the xy plane. The wire carries a current of 2.50 A. What is the torque on the wire if

there is a uniform magnetic field of 0.300 T directed

a. In the x direction?

= N B A I sin () = (1)(0.300 T)(0.0600 m)2 (2.50 A) sin (

) = 0.0027 N-m

b. In the y direction?

= N B A I sin () = (1)(0.300 T)(0.0600 m)2 (2.50 A) sin (

) = 0.0027 N-m

c. In the z direction?

= N B A I sin () = (1)(0.300 T)(0.0600 m)2 (2.50 A) sin (

) = 0 N-m

9. Three infinite wires, each carrying a current of 5.00 A, are shown below. What are the

magnitude and direction of the magnetic field at the point x = 2.00 cm, y = 6.00 cm

produced by these currents?

B1 = 0 I 1/(2d1)= (4x 10

-7 Tm/a)(5.00 A)/ ((2)(0.06 m)) = 0.167 x 10

-4 T, out of the page

B2 = 0 I 2/(2d2)= (4x 10-7

Tm/a)(5.00 A)/ ((2)(0.04 m)) = 0.250 x 10-4

T, out of the page

B3 = 0 I 3/(2d3)= (4x 10-7

Tm/a)(5.00 A)/ ((2)(0.02 m)) = 0.500 x 10-4

T, out of the page

Btot = B1 + B2 + B3 = 0.917 x 10-4

T, out of the page

You place a positive charge of 0.245 C at this location. What is the magnitude of a

magnetic force acting on the charge?

F = 0 since V = 0

10. When t = 3.00 s, the current in 60.0 mH inductor is 120 m A and is increasing at a rate of

25.0 mA/ms.

a. What is the magnitude of the emf induced in the inductor?

= - L t

= (0.0600 H) 25.0 A/s) = 1.50 V

b. What is the polarity of the induced emf?

Opposite to the polarity of the battery maintaining the current.

c. How much energy is stored in the inductor at that specific moment?

EST = ½ L I2 = ½ (0.0600 H) (0.120 A)

2 = 4.32 x 10

-4 J


NAME_____SOLUTION________________________












multiple choice and three (3) calculation problems. Show all work; partial credit will be given

for correct work shown.



7. Start: 10:30 a.m.

Stop: 11:20 a.m.

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

10

25

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. The force on a current-carrying wire in a magnetic field is the strongest when

a. The current is parallel to the field lines.

b. The current is at a 30° angle with respect to the field lines.

(5)

c. The current is at a 60° angle with respect to the field lines.

d. The current is perpendicular to the field lines.

2. A particle carrying a charge of +e travels in a circular path in a uniform magnetic field. If

instead the particle carried a charge of +2e, the radius of the circular path would have been

a. Twice the original radius.

b. Four times the original radius.

(5)

c. One-half the original radius.

d. One-fourth the original radius.

3. A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its

north pole pointing down, and released. As it approaches the coil, the falling magnet induces

(when viewed from above)

a. No current in the coil.

b. A clockwise current in the coil.

(5)

c. A counterclockwise current in the coil.

d. A current whose direction cannot be determined from the information provided.

4. All of the following have the same units except:

a. Inductance.

b. Capacitive reactance.

(5)

c. Impedance.

d. Resistance.

5. As the frequency of the AC voltage across an inductor approaches zero, the inductive

reactance of that coil

a. Approaches zero.

b. Approaches infinity.

(5)

c. Approaches unity.

d. None of the given answers

6. Resonance in a series RLC circuit occurs when

a. XL is greater than XC.

b. XC is greater than XL.

(5)

c. (XL - XC)2 is equal to R

2.

d. XC equals XL.

7. A 135-mH inductor with -0.2 resistance is connected in series to a F-20 capacitor and a

60-Hz, 45-V source. Calculate

a. The impedance of the circuit.

XL = 2 f C = 2 (60.0 Hz) (135 x 10-3

H) = 50.9

XC = 1/(2 f C) = 1/(2 (60.0 Hz) (20.0 x 10-6

F)) = 133

Z = (R2 + (XL – XC)

2)1/2

= 82.1

b. The rms current.

Irms = Vrms/Z = (45.0 V)/(82.1 ) = 0.548 A

c. The phase angle.

tan = (XL – XC)/R = (50.9 133 )/ (2.00

= -88.6o

d. The average power dissipated in the circuit.

P = Irms V rms cos = 0.602 W

8. An electron experiences the greatest force as it travels sm109.2 6 in a magnetic field when it

is moving northward. The force is upward and of magnitude N.102.7 13 What are the

magnitude and direction of the magnetic field?

Fmax = q V B

B = Fmax/ (q V) = (7.20 x 10-13

N)/[(1.60 x 10-19

C)(2.90 x 106 m/s)] = 1.55 T, EASTWARD

9. A single loop of a copper wire in a shape of a square 4.00 cm on each side is lying flat on

a horizontal table. A large electromagnet is positioned with its north pole above and to

the left a little so that the uniform magnetic field is downward onto a loop, making an

angle of 30o with the vertical.

a. Compute the average induced emf across the loop as the field varies linearly from 0 to its

final value of 0.500 T in 200 ms.

= N t

i = 0

f = B f A cos () = (0.500 T)(0.0400 m)2

cos(30o) = 6.93 x 10

-4 Wb

= (1)( 6.93 x 10-4

Wb – 0 Wb)/(0.200 s) = 0.00346 V

b. What is the direction of the induced current?

Counterclockwise, looking from above.

10. A circular coil 16.0 cm in diameter and containing nine loops lies flat on the ground. The

Earth’s magnetic field at this location has magnitude T1050.5 5 and points into the Earth

at an angle of 56.0° below a line pointing due north. If a 7.20-A clockwise current passes

through the coil, determine the torque on the coil.

= N B A I sin () = (9)(5.50 x 10-5

T)( (0.0800 m)2

(7.20 A)sin (34.0°) = 4.01 x 10-5

N-m


NAME___SOLUTION__________________________



Version 1 October 31, 2007









multiple choice problems and three (3) calculation problems. Show all work; partial credit will

be given for correct work shown.



7. Start: 11:30 a.m.

Stop: 12:20 p.m.

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

10

25

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. If the strength of the magnetic field applied to a loop of wire is doubled, what happens to the

magnetic flux through that loop assuming that all the other parameters remain unchanged?

(A) It stays the same.

(B) It is doubled.

(5) (C) It is tripled.

(D) It is quadrupled.

(E) It is reduced by a factor of 2.

2. Which of the following can be used for the SI-units of self-inductance?

(A) -s.

(B) H.

(5) (C) (J-s)/(C-A).

(D) All are correct answers.

(E) None is correct answer.

3. A proton, moving west, enters a magnetic field of a certain strength. Because of this field the

protons curves upward. What is the direction of the magnetic field?

(A) South.

(B) North.

(5)

(C) East.

(D) West.

(E) Downward.

4. Two long, parallel wires carry currents of different magnitudes. If the current in each wire is

doubled, what happens to the magnitude of the force between these two wires?

(A) It is doubled.

(B) It is tripled.

(5)

(C) It is reduced by a factor of 2.

(D) It stays the same.

(E) It is quadrupled.

5. A horizontal wire carries a current straight away from you. From your point of view, the

magnetic field caused by this current

(A) Points directly away from you.

(B) Points to the left.

(5)

(C) Circles the wire in a clockwise direction.

(D) Circles the wire in a counter-clockwise direction.

(E) Points to the right perpendicular to the wire.

6. A current in a solenoid coil creates a magnetic field inside that coil. The field strength is

directly proportional to:

(A) The coil area.

(B)The current.

(5)

(C) Both a and b are valid choices.

(D) None of the above choices is valid.

7. A proton of mass m = 1.67 x 10-27

kg and charge e = 1.6x 10-19

C moves in a circle of

radius 21.0 cm perpendicular to the magnetic field B = 0.400 T. Find

(a) Speed of the proton.

r = (mV)/(qB)

V = (r q B)/m = (0.210 m)( 1.6x 10-19

C)(0.400 T)/( 1.67 x 10-27

kg) = 8.04 x 106 m/s

(b) Period of the motion.

T = (2 r)/V = (2 (0.210 m))/(8.04 x 106 m/s) = 1.64 x 10

-7 s

8. Three infinite wires, each carrying a current of 5.00 A, are shown below. What are the

magnitude and direction of the magnetic field at the point x = 0 cm, y = 6.00 cm produced

by these currents?

B1 0 I1/(2 d1) = (4 x 10-7

Tm/A)(5.00 A)/( 2 (0.0600 m)) = 1.67 x 10-5

T, out of the page

B2 0 I2/(2 d2) = (4 x 10-7

Tm/A)(5.00 A)/( 2 (0.0400 m)) = 2.50 x 10-5

T, out of the page

B3 0 I3/(2 d3) = (4 x 10-7

Tm/A)(5.00 A)/( 2 (0.0200 m)) = 5.00 x 10-5

T, out of the page

Btot B1 B2 B3 9.17 x 10-5

T, out of the page

You place a stationary positive charge of 0.245 C at this location. What is the magnitude

of a magnetic force acting on the charge?

Stationary charge does not experience a magnetic force.

9. When t = 3.00 s, the current in 60.0 mH inductor is 120 m A and is increasing at a rate of 25.0

mA/ms.

d. What is the magnitude of the emf induced in the inductor?

= L t = (60.0 x 10-3

H)(25.0 A/s) = 1.50 V

e. How much energy is stored in the inductor at that specific moment?

U = ½ L I2 = ½ (60.0 x 10

-3H) (0.120 A)

2 = 4.32 x 10

-4 J

10. A rectangular loop consists of 100 closely wrapped turns and has dimensions 0.400 m by

0.300 m. The loop is hinged along the y axis and the plane of the coil makes an angle of 30.0o

with the x axis.

a.What is the magnitude of the torque exerted on the loop by a uniform magnetic field of 0.800 T

directed along the x axis, when the current in the loop has a value of 1.200 A in the direction

shown?

= N B A I sin () = (100)(0.800 T)((0.400 m) (0.300 m))(1.20 A)sin (60.0°) = 9.98

N-m

b. What is the expected direction of rotation of the loop?

Clockwise looking from above.

TIME OF COMPLETION_______________ NAME_____________________________



Version 1 March 30, 2006








5. There are six (6) multiple choice and four (4) calculation problems. Work all problems.

Show all work; partial credit will be given for correct work shown.



7. Start: 10:30 a.m.

Stop: 11:45 a.m.

PROBLEM

POINTS

CREDIT

1-6

30

7

15

8

20

9

20

10

15

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. A horizontal wire carries a current straight toward you. From your point of view, the


a. Points directly away from you.

b. Points to the left.

(5)

c. Circles the wire in a clockwise direction.

d. Circles the wire in a counter-clockwise direction.

2. The direction of the force on a current-carrying wire in a magnetic field is described by which

of the following?

a. Perpendicular to the current only.

b. Perpendicular to the magnetic field only.

(5)

c. Perpendicular to both the current and the magnetic field.

d. Perpendicular to neither the current or the magnetic field.

3. A charged particle is observed traveling in a circular path in a uniform magnetic field. If the

particle had been traveling twice as fast, the radius of the circular path would be

a. Twice the original radius.

b. Four times the original radius.

(5)

c. One-half the original radius.

d. One-fourth the original radius.

4. At double the distance from a long current-carrying wire, the strength of the magnetic field

produced by that wire decreases to

a. 1/8 of its original value.

b. 1/4 of its original value.

(5)

c. 1/2 of its original value.

d. None of the given answers.

5. A coil lies flat on a table top in a region where the magnetic field vector points straight up.

The magnetic field vanishes suddenly. When viewed from above, what is the direction of the

induced current in this coil as the field fades?

a. The induced current flows counterclockwise.

b. The induced current flows clockwise.

(5)

c. There is no induced current in this coil.

d. The current flows clockwise initially, and then it flows counterclockwise before

stopping.

6. A pure inductor is connected to an AC power supply. In this circuit, the current

a. Leads the voltage by 90°.

b. Lags the voltage by 90°.

(5)

c. Is in phase with the voltage.


7. The magnetic field perpendicular to a circular wire loop 12.0 cm in diameter is changed

from T52.0 to T45.0 in 180 ms, where means the field points away from an observer

towardand the observer.

a. Calculate the induced emf.

= -N t

i = Bi Ai = Bi ( ri2

) = 0.00588 Wb

f = Bf Af = Bf ( rf2

) = - 0.00509 Wb

= - (1) (-0.00509 Wb - 0.00588 Wb)/(0.180 s) = 0.0609 V

b. In what direction does the induced current flow?

Clockwise.

8. Two parallel wires each carry a current of 2.20 A in the same direction as shown below. The

distance between wires is 10.0 cm. A charge of 0.200 C moves parallel to the wires 15.0 cm to

the right from wire 2. The speed of the charge is 200 m/s.

a. Find the direction and the magnitude of the net magnetic field at the location of the charge.

B1 = 0 I1/(2d1) = (4 x 10-7

Tm/A)(2.20 A)/ /(2(0.250 m)) = 1.76 x 10-6

T, directed into the

page

B2 = 0 I2/(2d2) = (4 x 10-7

Tm/A)(2.20 A)/ /(2(0.150 m)) = 2.93 x 10-6

T, directed into the

page

Btot = B1 + B2 = (1.76 x 10-6

T) + (2.93 x 10-6

T) = 4.69 x 10-6

T, into the page

b. Find the direction and the magnitude of the magnetic force, which acts on the moving

charge.

F = B V q sin() = (4.69 x 10-6

T) (200 m/s)(0.200 C)(1) = 1.88 x 10-4

N, to the left

9. A 550-turn solenoid is 15.0 cm long. The current in it is 33.0 A. A 3.00-cm-long straight

wire cuts through the center of the solenoid, along a diameter. This wire carries a 22.0-A

current downward (and is connected by other wires that don’t concern us). What is the force

on this wire (specify both magnitude and direction) assuming the solenoid’s field points due

east?

B = 0 n I = (4 x 10-7

Tm/A)(550/(0.150 m))(33.0 A) = 0.152 T

F = B I l = (0.152 T)(22.0 A)(0.0300 m) = 0.100 N, out of the page (assuming that B

points to the right of the page and I flows down the page).

10. A coil with 150 turns, a radius of 5.00 cm, and a resistance of 12 surrounds

a solenoid with cmturns230 and a radius of 4.50 cm. The current in the solenoid

changes at a constant rate from 0 to 2.0 A in 0.10 s. Calculate the magnitude and

direction of the induced current in the coil.

(15)

= -N t

i = Bi Ai = 0

f = Bf Af = (0 n If) ( r2

) = (4 x 10-7

Tm/A)(23000)(2.00 A) ( (0.0450 m)2

) = 3.67 x

10-4

Wb

= - (150) (3.67 x 10-4

Wb - 0)/(0.100 s) = - 0.552 V

I = ||/R = (0.552 V)/(12.0 ) = 0.0460 A clockwise looking from the right, opposite to the

direction of current in the solenoid.


NAME___SOLUTION__________________________















7. Start: 10:30 a.m.

Stop: 11:45 a.m.

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9 20 10

20

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. A pure capacitor is connected to an AC power supply. In this circuit, the current

a. Leads the voltage by 90°.

b. Lags the voltage by 90°.

(4)

c. Is in phase with the voltage.


2. A bar magnet falls through a loop of wire with the north pole entering first. As the north pole

enters the wire, the induced current will be (as viewed from above)

a. Zero.

b. Clockwise.

(4)

c. Counterclockwise.

d. To top of loop.

3. A vertical wire carries a current straight up in a region where the magnetic field vector points

due north. What is the direction of the resulting force on this current?

a. Down.

b. North.

(4)

c. East.

d. West.

4. At a particular instant, a proton moves eastward at speed V in a uniform magnetic field that is

directed straight downward. The magnetic force that acts on it is

a. Zero.

b. Directed upward.

(4)

c. Directed to the north.

d. Directed to the south.

5. A long straight wire carries current toward the east. A proton moves toward the east

alongside and just south of the wire. What is the direction of the force on the proton?

a. North.

b. South.

(4)

c. Up.

d. Down.

6. Two long parallel wires are placed side-by-side on a horizontal table. If the wires carry

current in the same direction,

a. One wire is lifted slightly as the other wire is forced against the table's surface.

b. Both wires are lifted slightly.

(4)

c. The wires attract each other.

d. The wires repel each other.

7. Part of a single rectangular loop of wire with dimensions shown in below is situated inside a

region of uniform magnetic field of 0.550 T. The total resistance of the loop is .230.0

Calculate the force required to pull the loop from the field (to the right) at a constant

velocity of .sm40.3 Neglect gravity.

= V B L = (3.40 m/s)(0.550 T)(0.350 m) = 0.655 V

I = R = (0.655 V)/(0.230 ) = 2.85 A

F = I B L = (2.85 A)(0.550 T)(0.350 m) = 0.548 N

What is the direction of the current that flows in the loop?

Clockwise.

8. A particle with a + 2.00 C charge and a kinetic energy of 0.0900 J is fired into a uniform

magnetic field of magnitude 0.100 T perpendicularly to the field. If the particle moves in a

circular path of radius 3.00 m, determine its mass.

R = (mv)/(qB) = p/(qB)

p = R q B

K = p2/(2m)

m = p2/(2K) = (R q B)

2/(2K)

m = [(3.00 m)(2.00 x 10-6

C)(0.100 T)]2/(2 x 0.0900 J) = 2.00 x 10

-12 kg

If the magnetic field is perpendicular to the page and points up, does the particle rotate clockwise

or counterclockwise?

Clockwise.

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 20.0 cm long. The solenoid has 100 turns.

a. If the current in the solenoid is 3.00 A, find the magnetic flux through the square loop.

B = 0 n I = (4 x 10-7

Tm/A)(100/(0.200 m))(3.00 A) = 0.00188 T

= B A = (0.00188 T) (0.0100 m )2 = 1.88 x 10

-7 Wb

b. If the current in the solenoid is reduced to zero in 3.00 s, find the magnitude of the

average induced emf in the loop.

i = 1.88 x 10-7

Wb

f = 0

= N t = 1 (1.88 x 10-7

Wb)/(3.00s) = 6.27 x 10-8

V

10. A long, straight wire carries a current of 5.00 A. At one instant, a proton, 1.00 cm from the

wire travels at 1.50 x 103 m/s parallel to the wire and in the opposite direction as the current.

a. What is the magnitude and direction of the magnetic field at the location of the proton?

B = 0 I/(2d) = (4 x 10-7

Tm/A)(5.00 A)/(2(0.0100 m)) = 1.00 x 10-4

T, given that the current

flows north, directed into the page on the right side of the wire

b. Find the magnitude and direction of the force that acts on the proton because of the magnetic

field produced by the wire.

F = q V B sin () = (1.60 x 10-19

C)(1.50 x 103 m/s)(1.00 x 10

-4 T) = 2.40 x 10

-20 N, giv


NAME___SOLUTION__________________________












multiple choice problems and four (4) calculation problems. Show all work; partial credit will be

given for correct work shown.



7. Start: 10:30 a.m.

Stop: 11:45 a.m.

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. A current carrying circular loop of wire lies flat on a table top. When viewed from above, the

current moves around the loop in a counterclockwise sense. What is the direction of the

magnetic field caused by this current, inside the loop? The magnetic field

A) Circles the loop in a clockwise direction.

B) Circles the loop in a counterclockwise direction.

(4)

C) Points straight up.

D) Points straight down.

2. An electron has an initial velocity to the south but is observed to curve upward as the result of

a magnetic field. The direction of the magnetic field is

A) To the west.

B) To the east.

(4)

C) Upward.

D) Downward.

3. Two long parallel wires placed side-by-side on a horizontal table carry identical current

straight toward you. From your point of view, the magnetic field at the point exactly between

the two wires

A) Points up.

B) Points down.

(4)

C) Points toward you.

D) Is zero.

4. Kirchhoff's junction rule is an example of

A) Conservation of energy.

B) Conservation of charge.

(4)

C) Conservation of momentum.

D) None of the given answers

5. A long straight wire lies on a horizontal table and carries an ever-increasing current

northward. Two coils of wire lie flat on the table, one on either side of the wire. When viewed

from above, the induced current circles

A) Clockwise in both coils.

B) Counterclockwise in both coils.

(4)

C) Clockwise in the east coil and counterclockwise in the west coil.

D) Counterclockwise in the east coil and clockwise in the west coil.

6. A horizontal rod (oriented in the north-south direction) is moved westward at constant

velocity through a magnetic field that points straight up. Make a statement concerning the

potential induced across the rod.

A) The north end of the rod is at higher potential than the south end.

B) The south end of the rod is at higher potential than the north end.

(4)

C) Both ends of the rod are at the same non-zero potential.

D) Both ends of the rod are at zero potential.

7. 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.

I1 + I3 = I2

(-2.00 V) + I1 (1.00 ) + 4.00 V + I2 (2.00) = 0

I3 (3.00 ) + 4.00 V + I2 (2.00) + I3 (1.00 ) - 6.00 V = 0

I1 = - 8/7 A

I2 = - 3/7 A

I3 = 5/7 A

8. A specimen is to be exposed to a controllable magnetic field and is therefore positioned inside

a 0.550-m-long narrow air-core solenoid of cross-sectional area 2.00 x 10-4

m2. The

solenoid has 10 turns per cm. To monitor the field, a small coil (connected to a voltmeter)

is wrapped around the outside of the solenoid . When the current in the solenoid is increased

from zero to 4.90 A in 5.00 ms, what will be the emf indused in the coil given that it

consists of 240 turns of fine wire?

0 n I

i = Bi A = 0

f = Bf A 0 n I f A =( 4 x 10-7

Tm/A)(1000 turns/m)(4.90 A)( 2.00 x 10-4

m2) = 1.23 x 10

-6

Wb

= N t

= (240)( 1.23 x 10-6

Wb – 0)/(0.00500 s) = 59.1 x 10-3

V = 59.1 mV

9. In an experiment with cosmic rays, a vertical beam of particles that have charge of magnitude

3e and mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.250 T

and is bent in a semicercle of diameter 95.0 cm, as shown below.

a. Find the speed of the particles.

R = (mV)/(qB)

V= q B R/m

V = 3 (1.60 x 10-19

C)(0.250 T)(0.475 m)/(12 x 1.67 x 10-27

kg) = 2.84 x 106 m/s

b. What is the sign of their charge?

Negative

10. The triangular loop of wire shown below carries a current I = 5.00 A in the direction shown.

The loop is in the uniform magnetic field that has a magnitude of 3.00 T and the same direction

as the current in side PQ of the loop.

a. Find the force exerted by the magnetic field on each side of the triangle. If the force is not

zero, specify its direction.

F = I B L sin

FPQ = (5.00 A) (3.00 T) (0.462 m) sin

FPR = (5.00 A) (3.00 T) (0.800 m) sin 9into the page

FQR = (5.00 A) (3.00 T) (0.924 m) sin 6out of the page

b. The loop is pivoted about an axis what lies along side PR. What is the net torque exerted

on the loop?

IME OF COMPLETION_______________ NAME_____________________________



Version 1 July 10, 2006












7. Start: 2:30 p.m.

Stop: 3:50 p.m.

PROBLEM

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9

20

10

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FOR PARTIAL CREDIT.

1. A coil lies flat on a table top in a region where the magnetic field vector points straight up.

The magnetic field vanishes suddenly. When viewed from above, what is the sense of the

induced current in this coil as the field fades?

a. The induced current flows counterclockwise.

b. The induced current flows clockwise.

(4)

c. There is no induced current in this coil.

d. The current flows clockwise initially, and then it flows counterclockwise before

stopping.


due north. What is the direction of the resulting force on this current?

a. Down.

b. North.

(4)

c. East.

d. West.

3. An electron has an initial velocity to the south but is observed to curve upward as the result of


a. To the west.

b. To the east.

(4)

c. Upward.

d. Downward.

4. A horizontal wire carries a current straight toward you. From your point of view, the


a. Points directly away from you.

b. Points to the left.

(4)

c. Circles the wire in a clockwise direction.

d. Circles the wire in a counter-clockwise direction.

5. A current carrying loop of wire lies flat on a table top. When viewed from above, the current

moves around the loop in a counterclockwise sense. What is the direction of the magnetic field

caused by this current, outside the loop? The magnetic field

a. Circles the loop in a clockwise direction.

b. Circles the loop in a counterclockwise direction.

(4)

c. Points straight up.

d. Points straight down.

6. A horizontal rod (oriented in the east-west direction) is moved northward at constant velocity

through a magnetic field that points straight down. Make a statement concerning the potential

induced across the rod.

a. The west end of the rod is at higher potential than the east end.

b. The east end of the rod is at higher potential than the west end.

(4)

c. The top surface of the rod is at higher potential than the bottom surface.

d. The bottom surface of the rod is at higher potential than the top surface.

7. A rectangular loop of wire lies in the same plane as a straight wire, as shown below. There is

a current of 2.5 A in both wires. Determine the magnitude and direction of the net force on the

loop.

(20)

FL = - FR

FT = 0 I1 I2 l/(2d1) = (4 x 10-7

Tm/A)(2.50 A) )(2.50 A)(0.100 m)/(2(0.0300 m)) = 41.70 x

10-7

N, toward straight ware

FB = 0 I1 I2 l/(2d2) = (4 x 10-7

Tm/A)(2.50 A) )(2.50 A)(0.100 m)/(2(0.0800 m)) = 15.6 x 10-7

N, away from straight wire

FTOT = FB + FT + FL + FR = 41.70 x 10-7

N - 15.6 x 10-7

N = 26.1 x 10-7

N, toward straight wire.

8. A circular loop in the plane of the paper lies in a 0.75-T magnetic field pointing into the

paper. If the loop’s diameter changes from 20.0 cm to 6.0 cm in 0.50 s,

a. what is the direction of the induced current,

Clockwise.

b. what is the magnitude of the average induced emf,

= N t

i = B Ai = B ( ri2

) = (0.750 T) ( (0.100 m)2

) = 0.0236 Wb

f = B Af = B ( rf2

) = (0.750 T)( (0.0300 m)2

) = 0.00212 Wb

= (1)(0.00212 Wb – 0.0236 Wb)/(0.500 s) = 0.0430 V

c. if the coil resistance is ,5.2 what is the average induced current?

I = R

I = (0.0430 V)/(2.50 ) = 0.0172 A

9. The rod moves with a speed of ,sm6.1 is 30.0 cm long, and has a resistance of .5.2 The

magnetic field is 0.350 T, and the resistance of the U-shaped conductor is 0.25 at a given

instant. Calculate (a) the induced emf, (b) the current in the U-shaped conductor, and (c) the

external force needed to keep the rod’s velocity constant at that instant.

Note: Disregard figure (b) please.

(a)

=V B l = (1.60 m/s)(0.350 T)(0.300 m) = 0.168 V

(b)

I = R

I = (0.168 V)/(25.0 2.50 ) = 0.00611 A

(c) F = I B l = (0.00611 A)(0.350 T)(0.300 m) = 6.41 x 10-4

N, to the west


resistor if the values of the emfs and resistances were known. Label and indicate your choices

for current directions. DO NOT SOLVE THE EQUATIONS.

I1 = I2 + I3

2 - I2 R1 – I1 R4 + 3 – I1 R5 = 0

1 - I3 R3 + I2 R1 – I3 R2 = 0


NAME__SOLUTION___________________________















7. Start: 1:30 p.m.

Stop: 2:45 p.m.

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



FOR PARTIAL CREDIT.

1. If the number of turns in a rectangular coil of wire that is rotating in a magnetic field is

doubled, what happens to the induced emf, assuming all the other variables remain the same?

a. It stays the same.

b. It is reduced by a factor of 2.

(4)

c. It is reduced by a factor of 4.

d. It is doubled.

2. A proton, moving north, enters a magnetic field of a certain strength. Because of this field the

protons curves downward. What is the direction of the magnetic field?

a. Down ward.

b. Upward.

(4)

c. East.

d. West.

3. Three particles travel through a region of space where the magnetic field is out of the page, as

shown below. The electric charge of each of the three particles is, respectively,

a. 1 is neutral, 2 is negative, and 3 is positive.

b. 1 is neutral, 2 is positive, and 3 is negative.

(4)

c. 1 is positive, 2 is neutral, and 3 is negative.

d. 1 is negative, 2 is neutral, and 3 is positive.

4. At a particular instant, a proton moves eastward at speed V in a uniform magnetic field that is

directed straight downward. The magnetic force that acts on it is

a. Zero.

b. Directed upward.

(4)

c. Directed to the north.

d. Directed to the south.

5. Two long, parallel wires carry currents of different magnitudes. If the current in one of them

is doubled and the current in the other one is halved, what happens to the magnitude of the force

between these two wires?

a. It is doubled.

b. It is tripled.

(4)

c. It is reduced by a factor of 2.

d. It stays the same.

6. A rectangular coil lies flat on a horizontal table. A bar magnet is held above the center of the

coil with its south pole pointing down. What is the direction of the induced current in the coil

(looking from above)?

a. Clockwise.

b. Counterclockwise.

(4)

c. There is no current in the coil.

d. Away from the north pole and toward the south pole.

7. Part of a single rectangular loop of wire with dimensions shown in below is situated inside a

region of uniform magnetic field of 0.770 T. The total resistance of the loop is 5.00 .

a. With what constant velocity do you need to pull the loop out of the field in order to

generate electrical power of 10.0 W in the loop? Neglect gravity.

(20)

P2/RNBVL

R

V2P RNBL

VP RNBL

VP R(NBL)

V(10.0 W) (5.00 )(1(0.770 T)(0.350 m)) = 26.2 m/s

b. What force do you need to apply to the loop in order to move it with this constant velocity?

= N B V L = (1)(0.770 T)(26.2 m/s)(0.350 m) = 7.06 V

I = /R = (7.06 V) / (5.00 ) = 1.41 A

F = IBL = (1.41 A)(0.770 T)(0.350 m) = 0.381 N

c. What is the direction of the current that flows in the loop?

clockwise

8. An electron from the Sun with a speed of 1.00 x 107 m/s enters the Earth’s magnetic field

high above the equator where the magnetic field is 4.00 x 10-7

T. The electron moves

nearly in a circle except for a small drift along the direction of the Earth’s magnetic field

that will take it toward the North Pole.

a. What is the radius of the circular motion?

R = (mV)/(qB)

R = ((9.11 x 10-31

kg)(1.00 x 107 m/s))/((1.60 x 10

-19 C)(4.00 x 10

-7 T)) = 1.42 x 10

2 m = 142 m

b. What is the radius of the circular motion near the North Pole where the magnetic field is

2.00 x 10-5

T?

R = (mV)/(qB)

R = ((9.11 x 10-31

kg)(1.00 x 107 m/s))/((1.60 x 10

-19 C)(2.00 x 10

-5 T)) = 2.85 m

10. A current-carrying wire is bent into the shape of a square of sides L = 6.00 cm and

placed in the xy plane. It carries a current of 2.50 A. What is the torque on the wire if

there is a uniform magnetic field of 0.300 T

c. In the z direction, and

= N B A I sin () = (1)(0.300 T)(0.0600 m)2(2.50 A) sin (

) = 0

d. In the x direction?

= N B A I sin () = (1)(0.300 T)(0.0600 m)2 (2.50 A) sin (

) = 27.0 x 10

-4 N-m

10. Given the network, write equations that would allow you to solve for the currents in

each resistor if the values of the emfs and resistances were known. Label and indicate

your choices for current directions. DO NOT SOLVE THE EQUATIONS.

I1 = I2 + I3

1- I2 R4 + 2 - I1 R3 + 3 – I1 R5 – I1 R1= 0

2 + I3 R2 – I2 R4 = 0


NAME__SOLUTION___________________________












multiple choice problems and four (4) calculation problems. Show all work; partial credit will be

given for correct work shown.



7. Start: 1:30 p.m.

Stop: 2:45 p.m.

PROBLEM

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CREDIT

1-6

20

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20

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20

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20

10

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TOTAL

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PERCENTAGE



FOR PARTIAL CREDIT.

1. A proton in a magnetic field does not experience a force. Which of the following statements is

correct with respect to the situation?

I. The proton may have a velocity of zero m/s.

II. The proton may be moving parallel to the direction of the magnetic field.

(A) Statement I only.

(B) Statement II only.

(4) (C) Neither statement I nor statement II.

(D) Both statement I and statement II.

2. An electron moving in the positive y direction, at right angles to a magnetic field, experiences

a magnetic force in the negative x direction. What is the direction of the magnetic field?

(A) It is in the negative x direction.

(B) It is in the positive x direction.

(4) (C) It is in the positive y direction.

(D) It is in the negative z direction.

(E) It is in the positive z direction.

3. If the number of turns in the rectangular coil of wire that is rotating in a magnetic field is

doubled, what happens to the induced emf, assuming all the other variables remain the same?

(A) It stays the same.

(B) It is reduced by a factor of 4.

(4) (C) It is reduced by a factor of 2.

(D) It is doubled.

(E) It is quadrupled.

4. A horizontal rod (oriented in the east-west direction) is moved northward at constant

velocity through a magnetic field that points straight down. Make a statement concerning the

potential induced across the rod.

(A) The west end of the rod is at higher potential than the east end.

(B) The east end of the rod is at higher potential than the west end.

(4)

(C) The top surface of the rod is at higher potential than the bottom surface.

(D) The bottom surface of the rod is at higher potential than the top surface.

(E) All points of the rod are at the same potential.

5. A circular coil lies flat on a horizontal table. A bar magnet is held above its center with its

south pole pointing down. The stationary magnet induces (when viewed from above)

(A) No current in the coil.

(B) A clockwise current in the coil.

(4)

(C) A counterclockwise current in the coil.

(D) A current whose direction cannot be determined from the information given.

6. Kirchhoff’s loop rule is a statement of

(A) The law of conservation of momentum.

(B) The law of conservation of charge.

(4) (C) The law of conservation of energy.

(D) The law of conservation of angular momentum.

(E) Newton’s second law.

8. A rod 30.0 cm long moves at 8.00 m/s in a plane perpendicular to the magnetic field of

500 G (1 G = 10-4

T). The velocity of the rod is perpendicular to its length.

a. Find the magnetic force on an electron in the rod.

F = |q| B v sin = (1.60 x 10-19

C)(500 x 10-4

T)(8.00 m/s) sin( 90.0o) = 6.40 x 10

-20 N

b. What is the potential difference between the ends of the rod?

V = v B l = (8.00 m/s)(500 x 10-4

T)(0.300 m) = 0.120 V

c. Find the magnitude of the electric field in the rod.

V = E d

E = V/ d = (0.120 V)/(0.300 m) = 0.400 V/m

8. Two long straight parallel wires are 15 cm apart. Wire A carries 2.0 A current. Wire B’s

current is 4.0 A in the same direction.

a. Determine the magnetic field magnitude due to wire A at the position of wire B.

BA = 0 IA/(2r) = (4 x 10-7

Tm/A)(2.00 A)/(2(0.150 m)) = 2.67 x 10-6

T

b. Determine the magnetic field due to wire B at the position of wire A.

BB = 0 IB /(2r) = (4 x 10-7

Tm/A)(4.00 A)/(2(0.150 m)) = 5.33 x 10-6

T

c. Are these two magnetic fields equal and opposite? Why or why not?

Since two wires carry different currents, we cannot expect the magnetic fields to be the same in

magnitude. The two magnetic fields do point in the opposite direction though (use the right

hand rule)

d. Determine the force on wire A due to wire B, and the force on wire B due to wire A.

Are these two forces equal and opposite? Why or why not?

FAB /L = 0 IA IB /(2r) = (4 x 10-7

Tm/A)(2.00 A)(4.00 A)/(2(0.150 m)) = 1.07 x 10-5

N,

toward wire A

FBA /L = 0 IB IA /(2r) = (4 x 10-7

Tm/A)(4.00 A) (2.00 A)/(2(0.150 m)) = 1.07 x 10-5

N,

toward wire B

Yes, the forces are equal in magnitude and opposite in direction just as they should be according

to the Second Newton’s Law.

9. At the equator, a 100 turn circular coil with a cross-sectional area of 300 cm2 and a

resistance of 15.0 is aligned with its plane perpendicular to the Earth’s magnetic field

of 0.7 G. As the coil is realigned with its plane now parallel to the magnetic field,

a. What is the average emf induced in the coil if it takes 20.0 ms to realign the coil?

= - N t

i = B Ai cos i = (0.700 x 10-4

T) (300 x 10-4

m2

) cos(0o) = 2.10 x 10

-6 Wb

f = B Af cos f = (0.700 x 10-4

T) (300 x 10-4

m2

) cos(90o) = 0

= - (100)(0 – 2.10 x 10-6

Wb)/(0.0200 s) = 0.0105 V

b. What is the current flowing through the coil?

I = / R = (0.0105 V)/(15.0 ) = 0.000700 A




I’ll let you

handle this one yourselves.

F COMPLETION_______________ NAME__SOLUTION___________________________





1. Check your examination for completeness prior to starting. There are a total of nine (9)

problems on six (6) pages.





5. There are six (6) multiple choice and three (3) calculation problems. Work five (5)

multiple choice and three (3) calculation problems. Show all work; partial credit will be given for

correct work shown.



7. Start: 12:00 p.m.

Stop: 12:50 p.m

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5. Which of the following is correct?

a. When a current carrying wire is in your right hand, thumb in the direction of the

magnetic field lines, your fingers point in the direction of the current.

b. When a current carrying wire is in your left hand, thumb in the direction of the

magnetic field lines, your fingers point in the direction of the current.

(5)

c. When a current carrying wire is in your right hand, thumb in the direction of the

current, your fingers point in the direction of the magnetic field lines.

d. When a current carrying wire is in your left hand, thumb in the direction of the

current, your fingers point in the direction of the magnetic field lines.

6. A charged particle moves across a constant magnetic field. The magnetic force on this

particle

a. Changes the particle's speed.

b. Causes the particle to accelerate.

(5)

c. Is in the direction of the particle's motion.

d. Changes the particle's speed causing the particle to accelerate.

7. The magnetic forces that two parallel wires with unequal currents flowing in the opposite

directions exert on each other are

a. Attractive and unequal in magnitude.

b. Repulsive and unequal in magnitude.

(5)

c. Attractive and equal in magnitude.

d. Repulsive and equal in magnitude.

PHYS 1112 Exam 2, Version 1

Fall 2004 57


due south. What is the direction of the resulting force on this current?

a. Down.

b. North.

(5)

c. East.

d. West.

5. An electron has an initial velocity to the north but is observed to curve upward as the result of


a. To the west.

b. To the east.

(5)

c. Upward.

d. Downward.

6. Kirchhoff’s loop rule is a statement of

a. The law of conservation of momentum.

b. The law of conservation of charge.

(5) c. The law of conservation of energy.

d. The law of conservation of angular momentum.


Fall 2004 58

7. 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 20.0 cm long. The solenoid has 100 turns. The current in the

solenoid is 30.0 A flowing counterclockwise.

e. Find the strength of the magnetic field inside the solenoid.

B = 0 n I = (4 x 10-7

T-m/A)(100 turns/0.200 m)(30.0 A) = 0.0188 T

f. What is the direction of the magnetic field inside the solenoid?

Upward (out of the page)

g. Find the torque on the wire loop due to the magnetic field of the solenoid if the loop

carries a current of 2.00 A in clockwise direction.

= N I A B sin ( = (1) (2.00 A) (0.0100 m)2

(0.0188 T) sin (0o) = 0





Fall 2004 59

1 – I1 R3– I1 R4 + 1 – I1 R2 = 0

2 – I3 R1 + 3 – I3 R5 = 0

I1 + I2 + I3 = 0

9. A uniform magnetic field of 0.500 T is directed to the north. At some instant, a particle with

charge +0.0200 C is moving with velocity 2.00 m/s in a direction 30.0o north of east.

a. What is the magnitude of the magnetic force on the particle?

F = |q| B V sin ( = (0.0200 x 10-6

C) (0.500 T) (2.00 m/s) sin (60o) = 0.0173 x 10

-6 N

b. What is the direction of the magnetic force?

Out of the page

c. In what direction should the particle move to experience no magnetic force?


Fall 2004 60

North or south (parallel or anti-parallel to the direction of the magnetic field)


Fall 2004 61

Documents

TIME OF COMPLETION - Clayton State University... (1)(0.300 t)(0.0600 m)2 ... -7 tm/a)(5.00 a)/ ((2 )(0.02 m)) = 0.500 x 10-4 t, ... time of completion_____ name_____ department of

TIME OF COMPLETION - Clayton State University... (1)(0.300 t)(0.0600 m)2 ... -7 tm/a)(5.00 a)/ ((2 )(0.02 m)) = 0.500 x 10-4 t, ... time of completion_ name_ department of