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Transcript
Clayton State University
Department of Natural Sciences
Aug 27, 2007
Physics 1112 – Quiz 1
Name ____SOLUTION_________________________________
1.
Electrons carry a
a. Positive charge.
b. Negative charge.
c. Neutral charge.
d. Variable charge.
2.
The units of Coulomb’s constant are?
a. Nm2/C2.
b. Nm.
c.
Nm/C2.
d. C2.
3.
Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes
of the charges are doubled, and their separation is halved, then what happens to
the electrical force between these charges?
a. It increases by a factor of 16.
b. It increases by a factor of 8.
c. It is doubled.
d. It remains the same.
Clayton State University
Department of Natural Sciences
August 31, 2007
Physics 1112 – Quiz 2
Name ___SOLUTION__________________________________
1.
Particles of charge 50.0 mC, - 60.0 mC and 70.0 mC are placed in a line as shown
below. The center one is 0.500 m from each of the others. Calculate the net force
acting on the central charge due to the other two.
q1 = +50.0 x 10-3 C
q2 = -60.0 x 10-3 C
q3 = +70.0 x 10-3 C
F12 = k | q1|| q2|/r122 = (8.99 x 109 N-m2/C2) (50.0 x 10-3 C) (60.0 x 10-3 C) / (0.500 m)2 = 1.08 x
108 N, to the left
F32 = k |q2 ||q3|/r232 = (8.99 x 109 N-m2/C2) (60.0 x 10-6 C) (70.0 x 10-6 C) / (0.500 m)2 = 1.51 x
108 N, to the right
Ftot = F12 + F32= - 1.08 x 108 N + 1.51 x 108 N = 4.30 x 107 N (to the right)
Clayton State University
Department of Natural Sciences
September 7, 2007
Physics 1112 – Quiz 3
Name ______SOLUTION_______________________________
1. Electric field lines near positive point charges
a. Circle clockwise.
b. Circle counter-clockwise.
c. Radiate inward.
d. Radiate outward.
2. The magnitude of the electric field shown
a. Increases to the right.
b. Increases down.
c. Decreases to the right.
d. Decreases down.
3. Can electric field lines intersect in free space?
a. Yes, but only at the midpoint between two equal like charges.
b. Yes, but only at the midpoint between a positive and a negative charge.
c. Yes, but only in vacuum.
d. No.
Clayton State University
Department of Natural Sciences
September 12, 2007
Physics 1112 – Quiz 4
Name ______SOLUTION_______________________________
Find the potential at point P for the rectangular group of charges shown below.
Q1 = -8.00 C, Q2 = 8.00 C, and Q3 = 12.0 C.
V1 = k q1/r1 = (8.99 x 109 N-m2/C2) (-8.00 x 10-6 C)/(0.403 m) = -178 x 103 V
V2 = k q2/r2 = (8.99 x 109 N-m2/C2) (8.00 x 10-6 C)/(0.200 m) = 360 x 103 V
V3 = k q3/r3 = (8.99 x 109 N-m2/C2) (12.0 x 10-6 C)/(0.350 m) = 308 x 103 V
Vtot = -178 x 103 V + 360 x 103 V + 308 x 103 V = 489 x 103 V= 4.89 x 105 V
You want to move another charge, Q4 = 2.00 C, from infinity to point P. How much
work will be required to assemble the combination of charges?
W = U = Q4 Vtot = (2.00 x 10-6 C)(4.89 x 105 V) = 0.978 J
Clayton State University
Department of Natural Sciences
September 21, 2007
Physics 1112 – Quiz 5
Name ___SOLUTION__________________________________
1. Calculate the resistance per unit length of a 22-gauge nichrome wire of radius 0.321
mm.
The resistivity of nichrome is 150 x 10-8 -m.
R
L
1.00m
 (150 x 10-8 -m)
= 4.63 
A
 (0.000321m) 2
2. If a potential difference of 10.0 V is maintained across a 1.00 m length of the nichrome
wire, what is the current in the wire?
V = I R
I
V
R
10.0V
 2.16 A
I = 4.63
Clayton State University
Department of Natural Sciences
October 8, 2007
Physics 1112 – Quiz 6
Name __SOLUTION___________________________________
1. Which of the following statements is correct?
a. Earth’s geographic north pole is the north pole of the Earth’s magnetic field.
b. Earth’s geographic south pole is the south pole of the Earth’s magnetic field.
c. The north pole of a magnet points towards the Earth’s geographic north pole.
d. The north pole of a magnet points towards the Earth’s geographic south pole.
2. Two south poles
a. Will always repel each other.
b. Will always attract each other.
c. Will sometimes repel and sometimes attract each other.
d. Will neither repel nor attract each other.
3. A negative charge moves inside a uniform magnetic field as shown. What is the
direction of the magnetic force acting on the charge?
a. Toward the top of the page.
b. Toward the bottom of the page.
c. Into the page.
d. Out of the page.
Clayton State University
Department of Natural Sciences
October 15, 2007
Physics 1112 – Quiz 7
Name _SOLUTION____________________________________
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.
a. Determine the magnetic force acting on the side of the loop closest to the straight
wire. (Do not forget to specify direction).
F = 0 I1 I2 L/(2d1) = (4 x 10-7 Tm/A)(2.50 A)(2.50 A)(0.100 m)/(2(0.0300 m)) = 4.17
x 10-6 N, toward straight wire
b. Determine the magnetic force acting on the side of the loop farthest from the
straight wire. (Do not forget to specify direction).
F = 0 I1 I2 l/(2d2) = (4 x 10-7 Tm/A)(2.50 A)(2.50 A)(0.100 m)/(2(0.0800 m)) = 1.56
x 10-6 N, away from straight wire
c. Determine the total magnetic force acting on the loop.
FTOT = 4.17 x 10-6 N – 1.56 x 10-6 N = 2.61 x 10-6 N, toward straight wire.
Clayton State University
Department of Natural Sciences
October 24, 2007
Physics 1112 – Quiz 8
Name ______SOLUTION_______________________________
Part of a single rectangular loop of wire with dimensions shown in below is situated
inside a region of uniform magnetic field of 0.880 T. The total resistance of the loop
is 10.0 .
a. With what constant velocity do you need to pull the loop out of the field in
order to generate electrical current of 10.0 A in the loop? Neglect gravity.
=vBL
=IR
I R= v B L
v = (I R)/ (B L)
v = (10.0 A) (10.0 )/[ (0.880 T) (0.350 m)] = 325 m/s
b. What force do you need to apply to the loop in order to move it with this
constant velocity?
F = B L I = (0.880 T) (0.350 m)(10.0 A) = 3.08 N
Clayton State University
Department of Natural Sciences
November 12, 2007
Physics 1112 – Quiz 9
Name _____SOLUTION________________________________
1. A resistor is connected to an AC supply. On this circuit, the current
a. Leads the voltage by 90o.
b. Lags the voltage by 90o.
c. Is in phase with the voltage.
d. Leads the voltage by 180o.
2. Which of the following statements is correct for a series RLC circuit?
a. The voltage across the capacitor leads the voltage across
the inductor by 90o.
b. The voltage across the inductor leads the voltage across the
capacitor by 90o.
c. The voltage across the inductor leads the voltage across the
resistor by 180o.
d. The voltage across the inductor leads the voltage across the
capacitor by 180o.
3. The capacitive reactance in an ac circuit changes by what factor when the
frequency is tripled?
a. 1/3.
b. 1/9.
c. 3.
d. 9.
Clayton State University
Department of Natural Sciences
November 14, 2007
Physics 1112 – Quiz 10
Name __SOLUTION___________________________________
A 150-mH inductor with 3.00  resistance is connected in series to a 45.0 F
capacitor and a 60-Hz, 45-V source. Calculate
a. The impedance of the circuit.
XL = 2  f C = 2 (60.0 Hz) (150 x 10-3 H) = 56.5 
XC = 1/(2  f C) = 1/(2 (60.0 Hz) (45.0 x 10-6 F)) = 58.9 
Z = (R2 + (XL – XC)2)1/2 = 3.84 
b. The rms current that flows in the inductor.
Irms = Vrms/Z = (45.0 V)/(3.84 ) = 11.7 A
c. The phase angle.
tan  = (XL – XC)/R = (56.5 58.9 )/ (3.00 
 = -38.7o
d. The average power dissipated in the circuit.
P = Irms V rms cos = 411 W
e. Does the voltage in the inductor lead the voltage in the capacitor or lags
behind?
Voltage in the inductor leads the voltage in the capacitor by 180o.
Clayton State University
Department of Natural Sciences
November 27, 2007
Physics 1112 – Quiz 11
Name _SOLUTION____________________________________
1. The angle of incidence is the angle between
a. The incident ray and the reflected ray.
b. The incident ray and the normal drawn at the point where this ray meets the
mirror.
c. The incident ray and the surface of the mirror.
d. The reflected ray and the surface of the mirror.
2. A ray of light strikes a plane mirror perpendicular to the mirror.
What is the angle of reflection?
a. 0o.
b. 90o.
c. 180o.
d.
360o.
3. A beam of light in air is incident on a glass slab, After entering the glass slab
a. The beam does not change its path.
b. The beam follows the normal to the glass slab.
c. The beam bends away from the normal.
d. The beam bends toward the normal.