Name _________________
Solutions to Test 2
March 19, 2008
This test consists of three parts. Please note that in parts II and III, you can skip one
question of those offered.
Possibly useful formulas:
Constants
e = 1.602 ×10−19 C
ε 0 = 8.854 ×10−12 C2 /N/m 2
μ0 = 4π ×10−7 T ⋅ m/A
RC Circuits
Q = Q0 e −t τ
Q = CE (1 − e
Circles
C = 2π R
A = π R2
−t τ
)
Cyclotro
n Motion
mv = qRB
qB
ω=
m
Loops
τ = IA × B
U = − IA ⋅ B
Triangles
A = 12 BH
Spheres
A = 4π R 2
V = 43 π R 3
Hall Effect
IB
ΔVH =
tnq
Field from Wire
μI
B = 0 ( cos θ1 + cos θ 2 )
4π a
μI
B= 0
2π a
Biot-Savart
μ I ds × rˆ
B= 0 ∫ 2
4π
r
Cylinders
V = π R2 L
Alat = 2π RL
Solenoid
Bin = μ0 NI A
Cones
V = π R2 L
1
3
Alat = π R L2 + R 2
Part I: Multiple Choice [20 points]
For each question, choose the best answer (2 points each)
1. If you have two resistors R1 and R2, and you wish to combine them to make a resistor
with less resistance than either one, you could connect them in
A) series, but not parallel
B) parallel, but not series
C) series and parallel will both work
D) neither series nor parallel will work
E) it is impossible to tell without knowing the specific values of the resistances
2. A superconducting (perfectly conducting) wire loop lies in a region with a magnetic
field passing through it. Suddenly, the wire loop is doubled in size, by stretching the
wires. How does the total magnetic flux through the wire loop change as it increases?
Include any effects caused by currents that begin to flow through the wire loop.
A) The flux gets twice as big
B) The flux gets four times as big
C) The flux gets half as big
D) The flux gets one-fourth as big
E) The flux stays the same (by Lenz’s Law)
3. Electric charges produce electric fields. What else can also produce electric fields?
A) Constant electrical currents
B) Moving magnetic charges
C) Constant magnetic fields
D) Changing magnetic flux
E) Moving neutral conductors
4. Which of the following describes the type of current that is produced in a commercial
generator and experienced when you plug something into an outlet
A) The potential difference is constant and always has the same sign
B) The potential difference is always the same sign, but varies widely in magnitude,
though it is never zero
C) The potential difference is always the same sign, though it can also be zero, but it
never reverses
D) The potential difference is always the same magnitude, but it switches from
positive to negative and back again many times a second
E) The potential difference changes in both magnitude and sign, varying
smoothly from a maximum (positive) to minimum (negative), and back again
5. In an ideal (infinitely long, closely spaced wire loops) solenoid, how does the
magnetic field inside the solenoid compare to the magnetic field outside the solenoid?
A) It is the same magnitude and points in the same direction
B) It is the same magnitude but points in the opposite direction
C) It is zero outside the solenoid, but not zero inside
D) It is zero inside the solenoid, but not zero outside
E) None of the above are correct
6. If two long, parallel wires have current flowing through them in opposite directions,
the wires will
A) Attract each other
B) Repel each other
C) Cause each wire to feel a stretching force, trying to make it longer
D) Cause each wire to have a compression force, trying to make it shorter
E) Parallel wires have no force on each other.
7. A loop of wire in a constant magnetic field carrying a current I can feel what kind of
forces?
A) Torque, but no net force
B) Net force, but no torque
C) Net force or torque
D) Neither net force nor torque
E) Compression forces, but not expansion forces
8. A ground fault circuit interrupter (GFCI or GFI) cuts off the current to an outlet
whenever
A) There is more than a specified amount of current flowing through the live wire
B) There is more than a specified amount of current flowing through the neutral wire
C) The current going back through the neutral wire doesn’t match the current
flowing out through the live wire
D) There is too little current flowing through the neutral wire
E) There is too little current flowing through the live wire
9. Calculate the integral
v∫ B ⋅ ds for the dashed loop sketched below.
Several current
values flowing through the plane of the paper are drawn in.
A) + ( 2 A ) μ0
6A
B) + ( 4 A ) μ0
3A
2A
C) − ( 2 A ) μ0
1A
D) − ( 4 A ) μ0
E) None of the above
10. Between the plates of a parallel plate capacitor that is being charged up, will there be
a magnetic field? Why or why not?
A) Yes, because actual electrical current is flowing between the two plates
B) Yes, because the changing electric field acts as a “displacement current” that
produces magnetic fields
C) Yes, but only if there is a dielectric material in between with magnetic properties
D) No, because there is no current actually flowing in this region
E) No, because the actual current flowing is cancelled by the displacement current
Part II: Short answer [20 points]
Choose two of the following questions and give a short answer (1-3 sentences) or
brief sketch (10 points each).
11. The capacitor at right contains charge Q in it. Explain
qualitatively what happens when you close the switch. Use
equations if you wish
The difference in potential across the two sides of the capacitor means there is
also a difference in potential across the resister, and current will flow, gradually
discharging the capacitor. If R is the resistance and C the capacitance, the charge will
diminish in a characteristic time τ = RC, and will decrease according to the formula
Q = Q0e −t τ .
12. Compare and contrast Gauss’s Law for electric and magnetic fields. What is
fundamentally different about magnetic fields that makes the rule different?
Gauss’s Law for electric fields says the total electric flux out of a region is
proportional to the total charge in that region. Gauss’s Law for magnetic fields says the
total magnetic flux out of a region is zero. In equations, we have
v∫ E ⋅ nˆ dA = q
in
S
ε0
and
v∫ B ⋅ nˆ dA = 0 .
S
The fact that we obtain zero for the magnetic flux indicates that, unlike electric fields,
there are no monopole sources of magnetic fields. It means you can’t have any “north”
magnetic poles without having “south” magnetic poles.
13 Explain qualitatively the Hall effect. It is not necessary to use any equations.
The Hall effect occurs when you have a current carrying “wire” in the presence of
a magnetic field. The charge inside the wire are pushed to one side of the wire by the
magnetic forces on them. They accumulate on one side of the wire, producing an electric
field which eventually cancels further flow of charge caused by the magnetic forces.
Thus a potential difference is set up between one side of the wire and the other.
3A
I2
I1
14. Sketched at right is part of a circuit. Not shown is the
+
rest of the circuit, which has 3 A flowing in along the
3 A – 12 V A
top wire, and 3 A flowing out along the bottom wire.
Call the currents I1 and I2 the two currents flowing in
each of the resistors, in the presumptive directions sketched.
(a) [5] Use Kirchoff’s first law to derive a relationship between the currents
8Ω
4Ω
Part III: Calculation: [60 points]
Choose three of the following four questions and
perform the indicated calculations (20 points each)
Using the top node and Kirchoff’s First Law, which says that the current into a
node must equal the current out, we have
I1 + 3 = I 2
(b) [5] Use Kirchoff’s second law to write a second expression relating the two
currents
We’ll go around the loop in a clockwise fashion. As marked, both resistors are
going with the stream of the current, and hence represent a decrease in potential, while
the batter represents an increase in potential, so we have
0 = 12 − 4 I1 − 8 I 2
(c) [5] Solve for the two currents I1 and I2. Did we orient the two currents
correctly; that is, is the actual current in the directions chosen, or have we
chosen them incorrectly?
Substituting in the first equation into the second, we see that
0 = 12 − 4 I1 − 8 ( I1 + 3) = 12 − 4 I1 − 8 I1 − 24 = −12 − 12 I1
so
12 I1 = −12
We divide by twelve and then substitute this into the other formula to give
I1 = −1
and
I 2 = I1 + 3 = −1 + 3 = 2
So the currents are I1 = -1 A and I2 = 2 A. Hence the current in the 8 Ω resistor is indeed,
from top to bottom, as marked, but the other resistor also has current from top to bottom,
the opposite of what is marked.
(d) [5] We wish to check that the current I2 flowing through the 8 Ω resistor was
calculated correctly. Redraw the partial circuit diagram, adding an
ammeter and voltmeter to check the voltage and current across the 8 Ω
resistor. What current and voltage should be registered across this resistor?
I have dotted in the ammeter and voltmeter. The Ammeter will measure 2 A, of course,
and since the resistor satisfies ΔV= IR, the voltmeter will register 16 V.
V
26.2 cm
15. Pions come in three types, π0 with charge zero, and π± with charge ± e. A pion
enters a region with constant magnetic field B = 0.600 T pointing out of the plane
of the paper, passes through a velocity selector, and then into a region with just
the constant magnetic field.
(a) [6] The velocity selector is designed to accept only pions
with velocity v = 5.00 × 107 m/s to the right. What is the
magnitude and direction of the electric field E required in
the velocity selector to make sure this works?
The Lorentz Force on a particle of charge q is given by
F = q ( E + v × B ) . If we want the particle to go
straight, the electric field must point opposite to
π
v × B . Since the particle is moving to the right,
and the magnetic field points out of the plane,
v × B points downwards and the electric field must
therefore point upwards. Since v and B are
perpendicular, the magnitude of v × B is just vB, so
B
Electric and
Magnetic fields
Magnetic
fields only
E = vB = ( 5.00 ×107 m/s ) ( 0.6 T ) = 3.00 ×107 V/m .
(b) [10] After passing through the velocity selector, the pion curves in a graceful
arc and hits the wall as sketched a distance 26.2 cm away from where it
entered. What is the mass of the pion?
This is simple cyclotron motion, for which we have mv = qRB . The radius of the
circle is R = 13.1 cm, so that
−19
qRB (1.602 ×10 C ) ( 0.131 m )( 0.600 T )
m=
=
= 2.52 ×10−28 kg .
7
v
5.00 ×10 m/s
(c) [4] Was the particle a π0, a π+, or a π-?
As argued in part (a), v × B points downwards, but the particle accelerates
upwards, so it must be negatively charges. It is a π-.
16. A certain cable consists of three parallel conductors, one a
very slender wire perpendicular to this page, the second a
coaxial cylinder of radius 2 cm, and the third another coaxial
cylinder of radius 4 cm. The wire at the center is carrying a
total current of 1 A flowing directly out of the paper. The
first cylinder has 7 A flowing directly into the paper. The
outer cylinder may carry additional current in either
direction. At right is sketched an end-on view of the cable.
4 cm
(a) [8] The magnetic field at point A, 1 cm from the axis of
the cylinder, is measured. What is its magnitude and direction?
C
B
A
2 cm
For all three parts of this problem, we will use Ampere’s Law, which since we
have no changing electric fields, yields v∫ B ⋅ ds = μ0 I in . Now, imagine drawing a loop
that passes through the point A. By the symmetry of the problem, the magnetic field
must point in circles around the wire, and must be constant in magnitude. Hence
v∫ B ⋅ ds = B v∫ ds = 2π rB = μ I
0 in
For this innermost loop, only the central wire is included, so Iin = 1 A, and we find
( 4π ×10 T ⋅ m/A ) (1 A ) = 20.0 μT
μI
B = 0 in =
2π r
2π (1×10−2 m )
−7
Since we calculated the current using the right-hand rule with our thumbs pointing up,
the magnetic field is calculated counterclockwise, and at the point A points straight up.
(b) [5] The magnetic field at point B, 3 cm from the axis of the cylinder, is
measured. What is its magnitude and direction?
The only difference for this part is that the radius has been increased to r = 3 cm,
and therefore the second inner cylinder carrying –7 A must be included. The total
included current is therefore –6 A, and we have
( 4π ×10 T ⋅ m/A ) ( -6 A ) = −40.0 μT
μI
B = 0 in =
2π r
2π ( 3 ×10−2 m )
−7
This time the minus sign means the magnetic field points clockwise, so it is 40.0 μT
pointing to the right.
(c) [7] At point C, 5 cm from the axis, it is found that there is no magnetic field.
What is the total current and its direction in the outer cylinder?
We can use the exact same formula as before, but we will only get a zero
magnetic field if the total current passing inside the 5 cm loop is zero. Since the wire
plus inner cylinder contribute 1 A + (-7 A) = -6 A, we must compensate with +6 A in the
outer cylinder, or 6 A coming out of the paper.
B = 4000t 2 − 30t
6 cm
where t is the time in seconds and B is the magnetic field in tesla.
(a) [9] Find the total magnetic flux through the loop as a function of t.
Magnetic flux is just area times magnetic field (at least, when the magnetic field
is perpendicular to the plane of the loop), so the flux is
Φ B = BA = 12 Bs 2 =
1
2
(.06 m )
2
( 4000t
2
− 30t ) T = ( 7.2t 2 − 0.054t ) Wb ,
Where t is in seconds and the flux is in Webers. However, if we want the total flux
through all 500 turns, we must multiply this by 500 to yield
Φ B ,tot = N Φ B = 500 ( 7.2t 2 − 0.054t ) Wb = ( 3600t 2 − 27t ) Wb
(b) [9] At time t = 0.0100 s, the voltage across the two external wires is measured.
What is the voltage difference?
This is a straightforward application of Faraday’s Law, which says that
E= −
d Φ B ,tot
dt
=−
d
3600t 2 − 27t ) Wb = ( −7200t + 27 ) V
(
dt
Plugging in t = 0.01 s, we have
E = −7200 ( 0.01) + 27 = −72 + 27 = −45 V .
So the voltage difference is 45 V.
(c) [2] In part (b), which wire is the positive wire?
We have calculated the flux upwards; therefore, the EMF we found is in a
counter-clockwise sense, which means we are measuring the difference between the
upper and lower wires. Hence the upper wire is -45 V above the lower wire, which is the
same as saying the lower wire is the positive wire.
6 cm
17. A coil of wire is in the shape of a right triangle with both legs having size s
= 6 cm and consisting of 500 turns of wire. It lies in a region where the
magnetic field in the direction out of the plane is changing according
to the formula