Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Multiple choice questions [120 points]
Answer all of the following questions. Read each question carefully. Fill the correct
bubble on your scantron sheet. Each correct answer is worth 4 points. Each question
has exactly one correct answer.
Questions 1 through 6 all refer to the same problem.
Consider the circuit below, consisting of a battery, switch, wire, and two other circuit
elements, denoted by boxes 1 and 2. The graphs below represent possible measurements
of the potential difference, V2=Va-Vb, across circuit element 2 versus time. Time t = 0
(vertical line) denotes the time the switch is moved. The horizontal line denotes V2=0.
For each of the following cases, choose the appropriate graph for the potential difference
measurement across element 2.
S1
S2
1
a
2
b
V2
t
A
B
C
D
E
Name: ______________________________________________________________ Total Points: _______
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Questions 1-3: The switch has been in position S2 for a long time. At time t=0, the
switch is moved to position S1.
1
Box 1 = resistor, Box 2 = capacitor
A. graph A
B.
C.
D.
E.
2
t
τ
graph B charge of the capacitor VC = Vbat (1 − e )
graph C
graph D
graph E
Box 1 = inductor, Box 2 = resistor
A. graph A
B.
C.
D.
E.
3
−
−
t
τ
graph B VR = IR = Vbat (1 − e )
graph C
graph D
graph E
Box 1 = capacitor, Box 2 = resistor
A.
B.
C.
D.
E.
graph A VR = IR = Vbat e
graph B
graph C
graph D
graph E
−
t
τ
. For t<0, VR=0 since there is no current.
Questions 4-6: The switch has been in position S1 for a long time. At time t=0, the
switch is moved to position S2.
4
Box 1 = resistor, Box 2 = capacitor
A. graph A
B. graph B
C. graph C
D. graph D
−
t
τ
E. graph E discharge of the capacitor VC = Vbat e . For t<0, VC=Vbat
5
Box 1 = inductor, Box 2 = resistor
A. graph A
B. graph B
C. graph C
D. graph D
E. graph E The inductor opposes the change of the current. It keeps the current
running for a little while: VR = IR = Vbat e
−
t
τ
. For t<0, VR=Vbat
2
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
6
Box 1 = capacitor, Box 2 = resistor
A. graph A
B. graph B
C. graph C
D. graph D The current flows in the opposite direction (the capacitor is being
−
t
discharged). VR = IR = −Vbat e τ . For t<0, the capacitor is fully charged and
there is no current.
E. graph E
Questions 7 through 9 all refer to the same problem.
Consider the circuit below. It contains a battery with emf Vb = 10.0 V, a switch S, a
resistor with R = 100 Ω, and an inductor L with negligible resistance. A voltmeter is in
parallel with the inductor (assume that the voltmeter has an infinite internal resistance).
The switch is closed at time t=0 s.
At time t 0 = 5.0 × 10 −3 s after the switch is closed, the voltmeter reads V=3.68 V.
S
R=100Ω
Vb=10V
7
L
V
What will the voltmeter read at time t1 = 10.0 × 10 −3 s ?
A. 7.36 V
B. 1.35 V V = Vb e
−
t
τ
Thus, V (t1 ) = Vb e
−
t1
τ
= Vb e
−
2 t0
τ
2
 V (t 0 ) 
3.68 
 = 10 × 
= Vb 

 10 
 Vb 
2
C. 1.84 V
D. 6.01 V
E. 6.32 V
8
What are the units for the value of the inductance
A.
B.
C.
D.
E.
tesla
weber
henry
ohms
farad
3
Name: ______________________________________________________________ Total Points: _______
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9
What is the magnitude of the value of the inductance (in the above units)?
−
t
A. 0.50 Use V = Vb e τ and τ =
B.
C.
D.
E.
L
, thus L = − R
R
t0
 V (t 0 ) 

ln 
 Vb 
5.0 × 10 −5
5.0 × 10 5
2.0 × 10 4
20
4
Name: ______________________________________________________________ Total Points: _______
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Questions 10 through 19 all refer to the same problem.
The picture below shows two current-carrying wires. A long wire parallel to the y-axis
crosses the x-axis at x = 10.0 cm and carries an upward current I1 = 2.0 A. A square loop
of side a = 10.0 cm lies in the xy-plane, centered at the origin. It carries a counterclockwise current I2 = 3.0 A. The +z direction points out of the page.
y
I1 = 2.0 A
10 cm
I2 = 3.0 A
10 cm
x
10 cm
Questions 10-12: What is the magnetic field vector (direction, units and magnitude)
along the right segment of the loop (at x = 5.0 cm) due to the vertical wire?
10 direction?
A. +x
B. +y
r
r µ 0 Idl × rˆ r
C. +z Use the right hand rule and dB =
dl is along +y, r̂ along -x
4π r 2
D. -x
E. -z
11 units?
A.
B.
C.
D.
E.
ampere
weber
volt
farad
tesla
12 magnitude (in the above units)?
5
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
A. 8.0 × 10 −6 Apply Ampere's law: Take as a closed path a circle in the xz plane
of radius 5cm and centered at x=10cm, z=0 cm. 2πrB = µ 0 I1 Thus
4π × 10 −7 × 2
2π × 0.05
×
2.5 10 −5
1.2 × 10 −5
4.0 × 10 −6
8.0 × 10 −8
B=
B.
C.
D.
E.
Questions 13-15:What is the force (direction, units and magnitude) on the right segment
of the loop (the portion with x = 5.0 cm, –5.0 cm < y < 5.0 cm) due to its interaction with
the vertical wire?
13 direction?
A.
B.
C.
D.
E.
r
r
r r
r
+x F = I 2 l × B and I 2 l is along +y, B along +z
+y
+z
-x
-z
14 units?
A.
B.
C.
D.
E.
newton
newton/meter
tesla
volt.meter
volt/meter
15 magnitude (in the above units)?
A.
B.
C.
D.
E.
1.6 × 10 −6
r r
r
2.4 × 10 −6 F = I 2 l × B, F = 3 × 0.1 × 8 × 10 −6
3.6 × 10 −6
6.0 × 10 −6
1.2 × 10 −5
16 The direction of the net force on the entire square loop is
A. +x From 13, the force on the wire at x=5cm is along +x. The force on the
wire at x=-5cm is along –x and is less than the force on the wire at x=+5cm,
6
Name: ______________________________________________________________ Total Points: _______
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(First)
since the wire is farther away. The forces on the wires at y=+5cm and y=5cm cancel out.
+y
+z
–x
no direction (F=0)
B.
C.
D.
E.
17 The direction of the net torque on the entire square loop is
A.
B.
C.
D.
E.
+z
+y
–z
–y
no direction (τ=0) All of the forces acting on the loop go through the center
of the loop.
Questions 18-19: The square loop is now rotated so that it lies in the yz-plane, centered
at the origin. It carries a counterclockwise current I2 = 3.0 A as viewed from the vertical
wire. The two vertical segments (at z = ± 5.0 cm) are equidistant from the vertical wire
(which is still at x = 10. cm, z = 0).
y
I1 = 2.0 A
10 cm
10 cm
x
I2 = 3.0 A
10 cm
z
18 The direction of the net force on the entire square loop is
A. +x
B. +y
C. +z No force on the wires at
y=+5cm and –5cm. For the wires
at z=+5cm and –5cm, see drawing
on the right.
Top view: y out of the page
F
B
x
F
B
z
7
Name: ______________________________________________________________ Total Points: _______
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(First)
D. –x
E. no direction (F=0)
19 The direction of the net torque on the entire square loop is
A.
B.
C.
D.
E.
+z
+y
–z
–y (see drawing above)
no direction (τ=0)
8
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Questions 20 through 30 all refer to the same problem.
Consider a square wire loop of side a lying in the xy plane, centered on the origin, as
shown below. The +z direction points out of the page in this drawing.
y
r
B
a
a
x
Questions 20-25:
r
A uniform magnetic field B = Bzˆ fills the region (field lines directed out of the page).
r
20 The magnitude of B is B=B0. What is the magnitude of the net flux through the loop?
A. B0 a 2 Φ=B0a2 taking the area vector along +z
B. 4aB0
C. 2 B0 a 2
D. πB0 a 2
π
E.
B0 a 2
4
Questions 20-25:The magnetic field magnitude is decreased linearly from B0 to 0 in a

t 
time t0, i.e. B = B0 1 − 
 t0 
Questions 21-22: At time t =
t0
, what is the induced emf (units and magnitude) around
2
the loop?
21 units?
A.
B.
C.
D.
E.
newton
joule
tesla
volt
ampere
9
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
22 magnitude?
A. B0 a 2 t 0
B.
B0 a 2
t0
C.
2 B0 a 2
t0
Φ = Ba 2 , emf = −
dΦ
dB B0 a 2
(independent of t)
= −a 2
=
dt
dt
t0
πB0 a 2
2t 0
E. Not enough information
D.
23 The direction of the current in the top segment of the loop (at y = +a /2) at time
t
t = 0 is
2
A. +x (clockwise in the loop)
B. –x (counterclockwise in the loop) To compute the flux, we chose the area
vector along +z (out of the page). The positive direction around the loop is
counterclockwise. Since the emf is positive, the current flows in a
counterclockwise direction.
C. no current in the loop
24 The direction of the net force on the square loop at time t =
A.
B.
C.
D.
E.
+x
+y
+z
-x
No direction (F=0) (see drawing)
t0
is
2
F
r
B
25 The direction of the net torque on the square loop at time t =
A.
B.
C.
D.
E.
F
t0
is
2
I
F
F
+x
+y
+z
–y
No direction (τ=0) All of the forces go through the center of the loop
10
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
Questions 26-30: The magnetic field is returned to its original value (B=B0). In a time t1,
the angle θ between the loop and the xy plane is increased from 0 to 60°, i.e.
t
θ (t ) = 60 o , and then held at θ = 60°. In the top view of the square loop below, +y
t1
points out of the page.
r
B = B0 zˆ
x
θ
z
n
26 At time t =
t1
(θ = 30°), the magnitude of the flux through the loop is
2
A. increasing
B. decreasing Φ=B0a2cosθ and cosθ decreases as θ goes from 0 to 60. (take the
area vector, n, as indicated on the figure)
C. not changing
27 The direction of the current in the top segment of the loop ( at y = +a /2) at time
t
t = 1 (θ = 30°) is (referring to the figure above)
2
A. towards the top right corner of the page
dΦ
> 0 . The positive
dt
direction around the loop is set by the choice of the area vector defined in 26.
C. no current in the loop
B. towards the bottom left corner of the page emf = −
28 The direction of the net force on the square loop at time t =
t1
(θ = 30°) is
2
A. +x
B. +y
C. +z
11
Name: ______________________________________________________________ Total Points: _______
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D. –x
E. no direction (F=0) Since B is uniform, the forces on any two opposite sides of
the square loop cancel out (same B, same length, opposite directions for the
current).
29 The direction of the net torque on the square loop at time t =
A.
B.
C.
D.
E.
t1
(θ = 30°) is
2
+x
+y τ = µ × B , µ is along n, B is along z
+z
–y
no direction (τ=0)
30 The magnitude of the induced emf in the loop is largest at
Α. θ=0°
Β. θ=30°
C. θ=45°
D. θ=60° emf = B0 a 2
dθ
π
sin θ = B0 a 2
sin θ .For θ between 0 and 60, sinθ is
dt
3t1
largest at θ=60.
E. all angles between 0° and 60° (the emf is constant)
12
Name: ______________________________________________________________ Total Points: _______
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PROBLEM 1[40 points]
Equipotentials in a plane perpendicular to 3 charged rods are shown in the figure below.
The rod on the left has charge +2q. The two rods on the right have charge –q. The line
connecting points BCD is a field line. The potential difference between adjacent
equipotentials is 1.0 V; the V=0 equipotential is labeled.
-q
+2q
-q
1). [10pts] On the figure, sketch the electric field line that passes through point A
(marked with black dot). Use an arrowhead to indicate direction.
The electric field line is perpendicular to the equipotential lines. The electric field is
directed from +2q to –q.
2). [10pts] Find the electric potential at points A, B, C and D. Explain.
Count the number of equipotential lines from the point to the V=0 equipotential line.
Use that ∆V=1V between two adjacent field lines. V decreases in the direction of the
electric field (which is from +2q to –q)
VA=+1V, VB=+5V, VC=+1V, VD=-1V
13
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
3). [10pts] Rank the magnitude of the electric field at points A, B, and C. Explain.
The smaller the spacing between the equipotential lines, the greater the electric field
∆V
since E =
∆r
The spacing between the equipotential lines is smallest at B, then at C then at A
EA<EC<EB
4). [10pts] How much work is done by the electric field in moving an electron from point
A to point D? Justify your answer. Is the work by the electric field positive or
negative? Explain.
W A→ D = q (V A − VD ) = − e(1 − (−1)) = −2 eV = −3.2 × 10 −19 J
The work done by the E field on the electron is negative.
14
Name: ______________________________________________________________ Total Points: _______
(Last)
(First)
PROBLEM 2[40 points]
In the circuits below, bulbs 1-5 are identical, and the batteries are identical and ideal.
Boxes X, Y, and Z contain unknown arrangements of bulbs.
It is known that Rx=Ry>Rz
1
+
_
2
X
+
_
4
3
Y
+
_
5
Z
1). [10 pts] Rank the potential difference across box X, box Y, and box Z in order of
increasing absolute value. Explain your reasoning.
Using Kirchhoff’s rule for the 3 circuits:
Vbat=VX+V1
Vbat=VY+V2
Vbat=VZ+V4
The total resistance of the circuit containing X is greater than the total resistance of the circuit
containing Y. This is because there is an extra path provided by bulb 3 in the circuit containing
Y and that RX=RY. Thus the current through bulb 1 (=current through the battery in the circuit
containing X) is greater than the current through bulb 2 (=current through the battery in the
circuit containing Y). It follows that V2>V1. Then using the first two equations above we find:
VX>VY
The total resistance of the circuit containing Z is less than the total resistance of the circuit
containing Y since RZ<RY. The current through bulb 4 is greater than the current through bulb
2. It follows that V4>V2. Then using to the last two equations above, we find VZ<VY.
To conclude:
VZ<VY<VX
2). [10 pts] Is the brightness of bulb 3 greater than, less than, or equal to the brightness of
bulb 5? Explain.
We have V3=VY and V5=VZ, thus V3>V5
The greater the voltage across a light bulb, the brighter the light bulb.
3 is brighter than 5.
15
Name: ______________________________________________________________ Total Points: _______
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3). Rank the brightness of bulbs 1, 2 and 4 in order of increasing brightness. Explain.
From the above discussion, V4<V2<V1
The greater the voltage across a light bulb, the brighter the light bulb. Thus in order of
increasing brightness:
1<2<4
4
4). [10pts] Suppose a second box identical to box
Z is connected to the circuit as shown.
Determine whether the brightness of each of
the bulbs below will increase, decrease, or
remain the same. Explain.
+
_
5
Z
Z
a. Bulb 4
Since the total resistance of the circuit decreases (one extra path), the
current through the battery increases. The current through 4 is the same as
the current through the battery. 4 gets brighter.
b. Bulb 5
V4+V5=Vbat
V4 increases since 4 gets brighter. Thus V5 decreases, and 5 gets dimmer.
16