TIME OF COMPLETION_______________            NAME__SOLUTION___________________________

 

 

                                           DEPARTMENT OF NATURAL SCIENCES

 

PHYS 1112, Exam 2                                                                                                             Section 1

Version 1                                                                                                                  October 28, 2002

Total Weight: 100 points

 

1.         Check your examination for completeness prior to starting.  There are a total of ten (10) problems on eight (8) pages.

 

2.         Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet  (provided by your instructor).

 

3.         You will have 80 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 all 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:                11:30 a.m.

Stop:                12:50 p.m

 

 

 

               PROBLEM

 

                 POINTS

 

                 CREDIT

 

                     1-6

 

                      30

 

 

 

                       7

 

                      20

 

 

 

8

 

                      15

 

 

 

                       9

 

                      15

 

 

 

                      10

 

                      20

 

 

 

                 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. The three loops of wire shown below are all in the region of space with a uniform, constant magnetic field. Loop 1 swings back and forth as the bob of pendulum, loop 2 rotates about a vertical axis, and loop 3 oscillates vertically on the end of a spring. Which loop or loops have a magnetic flux that changes with time?

 

    1. Loop 1.

 

    1. Loop 2.

(5)

    1. Loop 3.

 

    1. All of them.

 

 

  1. If you want to increase the magnetic field inside the solenoid, which of the following ways are better?

 

    1. Double the length of the solenoid, keeping the number of loops the same.

 

    1. Double the number of loops, keeping the length of the solenoid the same.

(5)

    1. Double the cross-sectional area of the solenoid, keeping the length the same.

 

    1. Decrease the current flowing inside the solenoid to the half of its initial value.

 

 

 

  1. A vertical length of copper wire moves across a horizontal magnetic field.

    1. No voltage is induced across its ends.

 

    1. A time-varying voltage is induced across its ends.

(5)

    1. Potential at the upper end is higher that the potential at the lower end.

 

    1. Potential at the lower end is higher that the potential at the upper end.

 

 

  1. An electron is moving inside the magnetic field as shown below. What is the direction of the magnetic force which acts on the electron?

 

    1. Into the page.

 

    1.  Out of the page.

(5)

    1. To the top of the page.

 

    1. No magnetic force acts on the electron.

 

 

 

  1. In an LCR circuit, when the inductive and capacitive reactances are equal,

 

    1. The resistance is zero.

 

    1. Voltage leads the current.

(5)

    1. Current is in phase with voltage.

 

    1. Current leads the voltage.

 

 

 

 

 

 

 

  1. The source of electromagnetic waves is:

 

    1. A constant current.

 

    1. Any accelerating charge.

(5)

    1. Any accelerating particle.

 

    1. None of these.

 

 

  1.  The resistor in the circuit shown below has a resistance of 12.0 W and consumes 5.00 W of power. The rod is 1.25 m long and moves to the left with a constant speed of 3.10 m/s.

 

    1. What is the strength of the magnetic field?

 

(20)       e = B L V                P =  e2 /R                       P =  I2 R

             P = (B L V)2 / R = (B2 L2 V2)/ R            

                       

             B2 = P R / (L2 V2) = (5.00 W) (12.00 W)/((1.25 m) 2 (3.10 m/s)2)

              B = 2.00 T

 

    1. What external force is required to maintain the rod’s constant speed?

 

P =  I2 R

I = (P / R)1/2 =  ((5.00 W)/ (12.00 W)) 2 = 0.645 A             

F = I L B sin(q) = (0.645 A) (1.25 m) (2.00 T) = 1.61 N  

 

    1. What is the direction of the current in the loop?

 

 

Clockwise

 

  1. A series L-C-R circuit contains a 0.800 mF capacitor, a 250 mH coil, a 25.0 Ohms resistor, and a generator producing a maximum voltage of 220 V at a frequency of 60 Hz. 

 

    1. Calculate the impedance of the circuit.

 

XL = 2 p f L = 2 p (60.0 Hz) (250 x 10-3 H) = 94.2 W

 

XC = 1/ (2 p f C) = 1/ (2 p (60.0 Hz) (0.800 mF)) = 3316 W

 

Z = ( R2  + (XL – XC)2)1/2 = ( (25.0 W)2  + (94.2 W – 3316 W)2)1/2 = 3222 W

 

    1. Calculate the rms voltage drop across the resistor, the capacitor, and the inductor.

 

V = Vmax/(2)1/2 = (220 V)/(2)1/2 = 156 V

 

I = V / Z = (156 V) / (3222 W) = 0.048 A

 

VL = XL I = (94.2 W)(0.048 A) = 4.54 V

VC = XC I = (3316 W)(0.048 A) = 159 V

VR = XR I = (25.0 W)(0.048 A) = 1.20 V

 

    1. Does the current in this circuit lead the voltage or does the voltage lead the current? EXPLAIN your answer.

 

tan (q) = (XL – XC )/ R = (94.2 W - 3316 W) / (25.0 W) = -129

q = -89.6o

Angle is negative hence current leads the voltage.

 

  1. A solenoid is 20.0 cm long, has 200 loops, and carries a current of 3.25 A. A 15.0-mC charged particle moves at 1050 m/s through the interior of the solenoid, at an angle of 11.5o relative to the solenoid’s axis.

 

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

 

 

 

B = m0  n I=  (4 p x 10-7 T m/A)(200 loops/ (0.200 m)) (3.25 A) = 0.00408 T

 

 

    1.  Find the magnitude of force exerted on the particle.

 

F = q V B sin(q) = (15.0 x 10-6 C) (1050 m/s) (0.00408 T) sin(11.5o) = 12.8 x 10-6 N

 

 

  1. Two long, straight wires are oriented perpendicular to the page, as shown below.                              The current in one wire is I1 = 3.00 A, pointing into the page, and the current in the other wire is I2 = 4.00 A, pointing out of the page. Find the magnitude and direction of the net magnetic field at point P.

 

 

B1 = m0 I1 / (2 p r) = (4 p x 10-7 T m/A)(3.00 A)/(2 p (0.05 m)) = 120 x 10-7 T

 

B2 = m0 I2 / (2 p r) = (4 p x 10-7 T m/A)(4.00 A)/(2 p (0.05 m)) = 160 x 10-7 T

 

B1x = B1 cos(0 o) =   120 x 10-7 T

B1y = B1 sin(0 o) =   0  T

 

B2x = B2 cos(-90 o) =  0 T

B2y = B2 sin(-90 o) =   - 160 x 10-7 T

 

Btot x = B1x  + B2x  = 120 x 10-7 T

 

Btot y = B1y  + B2y  = - 160 x 10-7 T

 

Btot  = 200 x 10-7 T

q = tan-1(Btot y / Btot x) = -53o