Problem 1: At 20 °C, copper has a resistivity ρ = 1.7×10−8 Ω·m and α = 0.0039/°C.
Part a (4 pts): When a 200 meter long length of copper wire is connected across a 12 V
potential difference, a current of 5 A flows through it. What is the diameter of the wire?
Lρ
V
R = πr
2 = I
r = 0.672 mm
d = 1.34 mm
Part b (4 pts): At what temperature will the resistance of the copper wire be 15%
larger than it is at 20°C?
ρ(T ) = ρ(T 0 )[1 + α(T − T 0 )]
1.15 = 1 + α(T − T 0 )
T = T 0 + 0.15/α
T = 58.46 degrees
Problem 2: A coil of length 2cm has a turn density of 250 turns/mm. The coil is
rectangular in shape with sides of length 0.1 m and 0.2 m. The center of the coil is
along the z-axis and the coil carries a current that is moving counterclockwise around
the axis. When the coil is in the presence of an external, uniform magnetic field of
strength 4 T it feels a torque of 5 Nm. The magnetic field is in the direction < a, b, 0 >
(Assume that this direction is a unit vector).
Part a (4 pts): What is the current in the coil?
~
τ = ~µ × B
µ = NIA
τ
I = NAB
I = 12.5 mA
Part b (4 pts): What is the direction of the torque vector?
< −b, a, 0 >
Part c (4 pts): If an electric dipole had a dipole moment of 2 × 10−29 Cm aligned along
the z-axis, how strong would an electric field in direction < a, b, 0 > have to be to
generate the same torque?
τ = ~p × E~
E = τ/p
E = 2.5 × 1029 N/C
Problem 3 (7 pts): Find the voltage across, current through and power dissipated
through each resistor in the following figure.
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The effective resistance of the 5 and 8 ohm resistors is 40/13. The total effective resistance of the circuit is 131/13 ohms. Since the voltage is 5 V the current through the
7 ohm resistor is 65/131=0.496 A. The 8 ohm resistor will have 5/13 of the current
which is 0.191 A and the 5 ohm resistor will get 8/13 of the current which is 0.305 A.
The voltages are given by V = IR so V7 = 3.47 V, V8 = 1.53 V and V5 = 1.53 V. The
powers are given by P = IV so P7 = 1.72 W, P8 = 0.291 W and P5 = 0.466 W.
Problem 4 (7 pts): In the following figure, the wire is carrying a current of 10 A in
from the bottom and out to the left. An external magnetic field of 250 µT is present and
directed into the page. Find the total force (magnitude and direction) felt by the wire.
The length of the vertical component of the wire is 0.5 m, the length of the horizontal
component of the wire is 1 m and the radius of the circular component is 0.5 m.
~
dF~ = Id~I × B
F~vert = (10A)(0.5m)(250µT)(−ˆi)
F~vert = −1.25 × 10−3 N ˆj
~ = (10A)(1m)(250µT)(− ˆj)
Fhoriz
~ = −2.5 × 10−3 N ˆj
Fhoriz
R
~ = −IB π/2 rdθ(cos θ xˆ + sin θˆy)
Fcurve
0
~ = −1.25 × 10−3 N(ˆi + ˆj)
Fcurve
~ = −2.5 × 10−3ˆi − 3.75 × 10−3 ˆj N
Ftotal
Problem 5 (7 pts): What is the effective capacitance of the following circuit?
2
Here the 1, 2 and 3 µF capacitors are one branch all in series with each other. The 7
and 8 µF capacitors are in parallel with each other while being in series with the 5 µF
capacitor.
1 −1
) + 4 = 8.3µF.
Ceff = ( 11 + 12 + 13 )−1 + ( 51 + 7+8
Problem 6 (7 pts): What is the effective resistance of the following circuit?
This is actually simply three resistors in parallel.
Re f f = R11 + R12 + R13 −1
Problem 7: In the following figure we have built a detector that uses a velocity filter to select particles with specific speeds.
Part a (2 pts each particle): Based on the paths provided, indicate which particles are
detected in which locations. The six different particles are protons (p), neutrons (n),
electrons (e− ), positrons (e+ ), He2+ and He1+ .
Path a=neutron
Path b=He+
Path c=He2+
Path d=proton
Path e=positron
Path f=electron
Part b (5 pts): If the magnetic field in the velocity selector and mass spectrometer
is 5 T, and the radius of the path of the protons is 1 m, what is the magnitude of the
electric field applied in the velocity filter?
3
r = mv
qB
v = E/B
mE
r = qB
2
2
E = rqB
m
E = 2.4 × 109 N/C
Problem 8: Use the following circuit for each part.
Part a (4 pts): Write down a system of four equations that you would use to solve this
circuit as a function of time using Kirchoff’s rules. Be sure to include at least one voltage loop rule and one junction rule. Use the current through R1 to be i1 , the current
through R2 to be i2 , the current through R3 to be i3 and the current through R to be i.
Keep a similar convention for the charge on the capacitors.
For parts b-d all the equations I am looking for use some combination of V, R, R1, R2,
R3, i1 , i2 and i3 but NOT i.
These are four examples I was thinking of:
V − i1 R1 − iR = 0
V − i2 R2 − q2 /C2 − iR = 0
V − i3 R3 − q3 /C3 − iR = 0
i = i1 + i2 + i3
Part b (3 pts): At the instant the switch is closed (t = 0), what is the effective resistance of the circuit?
Re f f = R + ( R11 + R12 + R13 )−1
Part c (3 pts): After a long time (t → ∞), what is the current through R1?
V
i1 (∞) = R+R
1
Part d (3 pts): After a long time (t → ∞), what is the charge on capacitor C3?
q3 (∞)
VR
C3 = V − R+R1
3 R1
q3 (∞) = VC
R+R1
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Problem 9: A device to measure an object’s weight has been set up using the following
picture. A mass is hung from the center of a bar that is in the presence of a 5 T field
directed into the plane of the figure. The battery voltage can be adjusted to change the
current through the circuit. The horizontal bar is 100 cm long and is very light-weight.
The thin vertical wires support no tension. All of the weight of the suspended mass is
supported by the magnetic force on the bar.
Part a (3 pts): Which side of the battery (left or right) should be the positive terminal?
Left in order to create the proper current.
Part b (4 pts): If the maximum terminal voltage of the battery is 400 V what is the
greatest mass the instrument can measure?
F = ILB = mg
LB
m = VRg
m = 40.77 kg
Problem 10: Part a (4 pts): Explain the physical reason why the effective capacitance
of capacitors in parallel is the sum of the individual capacitances.
A capacitor is a device for storing charge. It stores charge inversely proportional to
the voltage applied across the capacitor and directly proportional to the capacitance. In
a parallel situation the same voltage is applied across many capacitors, in which case
they store a charge independent of each other. Thus looking like one large capacitor
with the charge being the total charge collected on a larger capacitor with an equivalent
voltage.
Part b (4 pts): Explain the physical reason why the effective resistance of resistors in
series is the sum of the individual resistances.
In a series circuit the current flows through each device in turn. We know that the
voltage drop across a resistor is the current multiplied by the resistance. In a series cir5
cuit the voltage drop across the entire series must then be the sum of the voltage drops
across each device. Since the current is the same in each resistor, the circuit appears to
be one larger resistor with a resistance equal to the sum of the series components.
Problem 11 (7 pts): What is the force vector of an electron traveling in the positive
x-direction at 2 × 104 m/s in the presence of an external magnetic field of < −5, 6, 4 >
T?
~
F~ = −qe~v × B
~v =< v x , 0, 0 >
~ =< −5, 6, 4 >
B
ˆ
F~ = 3.2 × 10−15 (4 ˆj − 6k)
F~ = 1.28 × 10−14 ˆj − 1.92 × 10−14 kˆ N
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