TIME OF COMPLETION_______________ NAME___SOLUTION_________________________

DEPARTMENT OF NATURAL SCIENCES

PHYS 1112, Exam 2 Section 1

Version 1 March 30, 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 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: 1:30 p.m.

Stop: 2:45 p.m

PROBLEM	POINTS	CREDIT
1-6	30
7	15
8	20
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. The unit of inductance is

1. H/s.

2. H.

(5)

3. W.

4. W-m.

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

1. To the right.

2. To the left.

(5)

3. Away from you.

4. Toward you.

3. A long, straight, horizontal wire located on the equator carries a current directed toward the east. What is the direction of the force on the wire due to the Earth’s magnetic field?

1. East.

(5)

2. West.

3. Upward.

4. Downward.

4. For an induced current to appear in a wire loop,

1. There must be a large magnetic flux through the loop.

2. The loop’s plane must be parallel to the magnetic field.

(5)

3. The loop’s plane must be perpendicular to the magnetic field.

4. The magnetic flux through the loop must vary with time.

5. Which of the following is true in an ac circuit?

1. V_max is greater than V_rms.

2. V_max is less than V_rms.

3. V_max is equal to V_rms.

4. Either one can be greater depending on frequency of the generator.

6. A capacitor is connected to a variable-frequency ac voltage source. If the frequency increases by the factor of 3, the capacitive reactance will be

1. 3 times the original reactance.

2. 1/3 of the original reactance.

(5)

3. 9 times the original reactance.

4. 1/9 of the original reactance.

7. A particle with a charge of –5.00 x 10^–4 C and a mass of 2.0 x 10^–9 kg moves at a speed of 1.00 x 10³ m/s in the +x direction toward a uniform magnetic field of 0.200 T in the +y direction.

1. Which way will the particle start to deflect as it enters the field?

Downward.

2. What is the magnitude of force on the particle as it enters the magnetic field?

F = q V B = (5.00 x 10^–4 C)(1.00 x 10³ m/s)(0.200 T) = 0.100 N

3. What is the radius of the circular orbit of the particle while it is the field?

r = (m V) / (q B)

r = [(2.0 x 10^–9 kg)( 1.00 x 10³ m/s)] / [(5.00 x 10^–4 C)( 0.200 T)] = 2.00 x 10^-2 m

8. A solenoid with 100 turns per centimeter carries a current of 1.20 A as shown below. Radius of the solenoid is 2.00 cm and its length is 10.0 cm.

n = 10,000 turns/m

a. What is the magnitude and the direction of the magnetic field inside the solenoid?

B_i = m₀ n I_i = (4 p x 10^-7 T m / A)(10,000 turns/m)(1.20 A) = 1.51 x 10^-2 T to the left

b. The current increases to 2.40 A in 100 ms. What is the magnitude and the direction of the magnetic field inside the solenoid now?

B_f = m₀ n I_f = (4 p x 10^-7 T m / A)(10,000 turns/m)(2.40 A) = 3.02 x 10^-2 T to the left

c. What is the change in the magnetic flux through the single turn of the solenoid?

DF = F_f – F_I = (B_f - B_i) A = (B_f - B_i)(p r²) = (1.51 x 10^-2 T)(p (0.0200 m)²) = 1.90 x 10^-5 Wb

d. What is the magnitude of the emf induced in the solenoid because of the change in the electric current?

e = N DF/Dt

e = (1000)(1.90 x 10^-5 Wb) / (100 x 10^-6 s) = 190 V

e. Bonus (worth 5 points): What is the direction of the induced emf?

The polarity of the induced emf is opposite to the polarity of the battery.

9. A circuit is connected to 110-V (Hint: this is a rms value), 60-Hz source contains a 50.0-W resistor and a coil with the inductance of 100 mH. .

1. What is the reactance of the coil?

X_L = 2p f L = 2p (60.0 Hz) (100 x 10^-3 H) = 37.7 W

2. What is the impedance of the circuit?

Z = (R² + (X_L – X_C)²)^1/2 = ((50.0 W)² + (37.7 W – 0)²)^1/2 = 62.6 W

3. What is the rms current in the circuit?

I_RMS = DV_RMS/Z

I_RMS = (110 V) / (62.6 W) = 1.76 A

4. How much power is dissipated by the coil?

None. Only resistor dissipates power in ac circuits.

5. Does the voltage lead the current in the circuit or the current leads the voltage? EXPLAIN.

f = tan^-1 ((X_L – X_C)/R) = 37.7^o

f = tan^-1 ((37.7 W – 0)/ (50.0 W)) = 37.7^o

f >0, hence voltage leads current.

10. A loop of current-carrying wire is in a 1.60-T magnetic field.

1. For the magnetic torque on the loop to be at maximum, should the plane of the coil to be parallel, perpendicular, or at 45^o angle to the magnetic field? EXPLAIN.

Plane of the coil should be parallel to the direction of the magnetic field, which means that the angle between the normal to the loop and the magnetic field becomes maximum (90^o) maximizing the torque.

2. If the loop is rectangular with dimensions 20.0 by 30.0 cm and carries the current of 1.50 A, what is the magnitude of the maximum torque exerted on the loop by the magnetic field?

t = N B A I = (1)(1.60 T)(0.200 m x 0.300 m)(1.50 A) = 0.144 N-m

3. If you change the shape of the loop from rectangular to circular keeping the total area of the loop the same, will the torque be greater, lesser, or the same?

Torque will not change, since it does not depend on the shape, but on total area.