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? 

 

    1. 28.2 A.

 

    1. 3.10 A.

(6)

    1. 7.07 A.

 

    1. 14.1 A.

 

 

 

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

 

    1. To the right.

 

    1. To the left.

(6)

    1. Away from you.

 

    1. Toward you.

 

 

 

  1. 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:

 

    1. Clockwise.

 

    1. Counterclockwise.

(6)

    1. Zero.

 

    1. Along the length of the magnet.

 

 

 

 

 

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

 

    1. Directly proportional to L2.

 

    1. Directly proportional to L1/2.

(6)

    1. Directly proportional to L.

 

    1. Inversely proportional to L.

 

 

 

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

 

    1. Straight line.

 

    1. Circle.

(6)

    1. Ellipse.

 

    1. Parabola.

 

 

 

 

  1. A current in a solenoid coil creates a magnetic field inside that coil. The field strength is directly proportional to:

 

    1. The coil area.

 

    1. The current.

(6)

    1.  Both a and b are valid choices.

 

    1. None of the above choices are valid.  

 

 

 

 

 

 

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

    1. to the right,

 

e = 0, since DF = 0

 

I = 0

    1. up,

 

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

 

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

 

 

    1. to the left,

 

e = 0, since DF = 0

 

I = 0

 

    1. down.

 

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

 

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

 

  1. 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?

 

t = N B A sin (q) = (1)(0.300 T)(0.0600 m)2 sin (90.0o) = 0.00108 N-m

 

b.     In the y direction?

 

t = N B A sin (q) = (1)(0.300 T)(0.0600 m)2 sin (90.0o) = 0.00108 N-m

 

c.      In the z direction?

t = N B A sin (q) = (1)(0.300 T)(0.0600 m)2 sin (0.0o) = 0 N-m

 

 

  1. 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 = m0 I 1/(2pd1)=  (4p x 10-7 Tm/a)(5.00 A)/ ((2p)(0.06 m)) = 0.167 x 10-4 T, out of the page

 

B2 = m0 I 2/(2pd2)=  (4p x 10-7 Tm/a)(5.00 A)/ ((2p)(0.04 m)) = 0.250 x 10-4 T, out of the page

 

B3 = m0 I 3/(2pd3)=  (4p x 10-7 Tm/a)(5.00 A)/ ((2p)(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

 

 

 

 

  1. 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. 

  

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

 

e = - L D I / D t

                                                                                                       

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

 

    1. What is the polarity of the induced emf?

 

Opposite to the polarity of the battery maintaining the current.

 

    1. 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