PHY2054 Fall 2014
Prof. Paul Avery
Prof. Andrey Korytov
Oct. 29, 2014
Exam 2 Solutions
1. A loop of wire carrying a current of 2.0 A is in the shape of a right triangle
with two equal sides, each with length L = 15 cm as shown in the figure. The
triangle lies within a 0.7 T uniform magnetic field in the plane of the triangle
and perpendicular to the hypotenuse. The magnetic force on either of the two
sides has a magnitude of:
B
Answer: 0.15 N
! ! !
Solution: The force on each side is F = I × B L,
(
)
F = IBL sin θ
F = BLi sin 45° = 0.7 × 0.15× 2.0 × 0.707 = 0.15 N.
2. Refer to the previous problem. What is the magnitude of the torque on this loop due to the
magnetic field (in N⋅m)?
Answer: 15.8 × 10−3
! ! !
Solution: The torque is τ = µ × B ,
(
(
)
τ = µ Bsin θ = (iA)Bsin θ
)
τ = iABsin 90! = 2.0 × 0.5× 0.152 × 0.7 = 1.58 ×10−2 N⋅m.
3. An electrical device with an equivalent resistance of 100 Ω has a maximum instantaneous
voltage rating of 60 V. Assume that it is plugged into a standard 120 V electrical outlet in series
with a resistance R. What is the approximate minimum value of R needed to keep the
instantaneous voltage across the device within its safety limits?
Answer: 183 Ω
Solution: The maximum instantaneous voltage from the 120 V outlet is 120 × 2 = 169.7 V. We
want to find the value of R such that the voltage across the device is no more than 60 V.
The device and resistor R are connected in series: RTOT=100+R. The current through the device
is I = 169.7/ RTOT. And the voltage across the device is V=100⋅I must satisfy the condition V≤60
V. Putting all together, we get 100 ⋅ 169.7 / 100 + R = 60 . Solving for R yields R = 183 Ω.
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4. A 14.0 g conducting rod of length 1.30 m and resistance 8.0 Ω slides freely
downward between two vertical conducting rails without friction. The entire
apparatus is placed in a 0.43 T uniform magnetic field. Ignore the electrical
resistance of the rails. What is the terminal velocity of the rod? (g = 9.8 m/s2)
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Answer: 3.51 m/s
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PHY2054 Fall 2014
Solution: The motional emf is ε = BLv which generates a loop current of i = ε / R = BLv / R .
At terminal velocity, the magnetic force on this current equals the force due to gravity or
mg = BiL = B 2 L2v / R . Solving yields v = 3.51 m/s.
5. A transformer consists of a 300-turn primary coil and a 2000-turn secondary coil. If the
current in the secondary coil is 9.00 A, what is the primary current?
Answer: 60.0 A
Solution: Using V2 / V1 = N 2 / N1 = 6.67 and i1V1 = i2V2 , with i2 = 9.0 A, we get i1 = 60.0 A.
6. A hair dryer has a power rating of 1200 W when plugged into a standard 110 V AC receptacle.
How much current does it draw if it is connected instead to a 110 V DC source?
Answer: 10.9 A
ε
ε /R,
from where one can find R. The current through a resistor connected to DC source emf ε is
P
I = ε / R . Therefore, I = ε
ε = 10.9 A.
Solution: Power dissipated into hit by a resistor connected to AC of rms emf
rms
is P =
2
rms
0
0
0
2
rms
7. The number of turns is tripled for an ideal solenoid, and its length is doubled while holding its
cross-sectional area constant. If the old inductance is L, what is the new inductance?
Answer: 9L/2
Solution: The inductance is proportional to the cross sectional area, the solenoid length and the
square of n = number of turns/length: L = µ0n 2 Aℓ
Then, if n → 3n / 2 (triple turns over twice the length) and l → 2l (length doubled), the new
(
)
2
(
)
inductance will be L old × 3 / 2 × 2 = 9 / 2 L old
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PHY2054 Fall 2014
8. In the circuit shown, what is the current through the 15 V
battery?
12 V
Answer: 1.18 A upwards
+
4Ω
Solution: We choose the currents in the branches to be I1 (left
branch upwards), I2 (middle branch upwards) and I3 = I1 + I2
(right branch downwards). Then the loop equations for leftmiddle and middle-right (others are possible, but these are convenient) are
+
15 V
8Ω
2Ω
−4I1 + 12 − 15 + 8I 2 = 0
−8I 2 + 15 − 2I3 = 0
Solving for I2 yields I 2 = +33 / 28  +1.18 A.
9. An ammeter with full-scale deflection for I = 20.0 A has an internal resistance of 31.5 Ω. We
need to use this ammeter to measure currents up to 30.0 A. What size resistor should be used to
protect the ammeter?
Answer: 63.0 Ω
Solution: The “shunt” resistance R is placed in parallel with the ammeter so that part of the
current to be measured is shunted around the device to avoid damaging it. We pick R so that at
30 A total current, 10 A goes through the shunt resistor and 20 A through the ammeter itself. The
voltage across the ammeter is 20 × 31.5 = 630 V, which is the same as the voltage across R. Thus
to get a current of 10 A through the resistor we require R = 63.0 Ω.
10. An electromagnetic flowmeter is used to measure blood flow rates
during surgery. Blood containing Na+ ions flow to the left through an
B
artery with an inner diameter of D = 4.00 mm. The artery is in a
downward magnetic field of B = 0.100 T and develops a Hall voltage
VH = 80.0 µV across its diameter. What is the volumetric flow rate of the blood in (m3/s)?
v
Answer: 2.51 × 10−6
Solution: In the Hall effect, ions are bent in the magnetic field toward the arterial walls until the
force FE = eE = e VH / D due to electric field E generated by the charges on opposite sides of
(
)
the diameter exactly balances the magnetic force FB = evB . The force equality condition is
e VH / D = evB . This yields v = VH / BD for the velocity. The volume flow rate F = area ×
(
)
(
)(
)
velocity or F = π D 2 / 4 VH / BD = π DVH / 4B = 2.51 × 10−6 m3/s.
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PHY2054 Fall 2014
11. An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has
1,700 turns, radius 2.4 cm, and length 13.0 cm. When 0.5 A of current flows through the
solenoid, the magnetic field inside the iron core has magnitude 0.43 T. What is the relative
permeability κB of the iron core?
Answer: 52.3
Solution: The B field of a solenoid with a ferromagnetic core is B = κ B µ0 ni , where κB is the
permeability of the core material. Here n = 1700 / 0.13 = 13,100 . Using the values shown we
obtain κB = 52.3.
12. Four long parallel wires pass through the corners of a square with side
14.1 cm as shown in the figure. All four wires carry the same magnitude of
current I = 5.7 A in the directions indicated. Find the magnetic field at the
center of the square.
Answer: 32.3 µT
Solution: Magnetic field lines make circles around lines of current, with
their direction going clockwise for wires with currents into the page and counterclockwise for
currents going out of the page. At the center of the square they point either at 45° (for the top-left
and bottom-right wires) or at -45° (for the top-right and bottom-left wires). By inspection, we see
that the y components of the field cancel at the center while the four x components are all
identical. If the side length is L, we obtain B = 4 × 2 µ0i / 2π L = 32.3 µT .
13. The armature of an AC generator is a rectangular coil 10.5 cm × 4.0 cm with 210 turns. It is
immersed in a uniform magnetic field of magnitude 1.5 T. If the amplitude of the emf in the coil
is 30.0 V, at what angular speed (in rad/sec) is the armature rotating?
Answer: 22.7
Solution: Emf induced in a coil of N loops and area A (A = ab ) rotating with angular velocity ω
in magnetic field B is ε = ω NBA⋅ sin ω t . Using the values shown for the emf amplitude, B field,
number of turns and coil dimensions yields ω = 22.7 rad/sec.
14. A dc motor has coils with resistance of 13 Ω and is connected to an emf of 127 V. When the
motor operates at full speed, the back emf is 63 V. If the current is 6.1 A with the motor
operating at less than full speed, what is the back emf at that time?
Answer: 47.7 V
ε
Solution: I = (
external
−
ε ) / R . Therefore, back emf is 127 − 13× 6.1 = 47.7 V.
back
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PHY2054 Fall 2014
15. A coil has an inductance of 0.31 H and a resistance of 43 Ω. The coil is connected to a 6.4 V
battery. After a long time elapses, the current in the coil is no longer changing. What is the
energy stored in the coil?
Answer: 3.43 mJ
Solution: After a long time the inductance has no effect and the current is obtained from
i = ε / R = 6.4 / 43 = 0.149 A. The energy stored in the coil is then U L = 12 Li 2 = 3.43 mJ .
16. A circuit breaker trips when the rms current exceeds 10.3 A. How many 60 W lightbulbs,
connected in parallel, can be run on this circuit without tripping the breaker? (The rms voltage of
the circuit is 110 V.)
Answer: 18
Solution: Each lightbulb draws a current of 60/110 = 0.545 A. Since 10.3/0.545 ~ 18.9, we see
that no more than 18 lightbulbs can be supported on this circuit.
17. A sample containing carbon (atomic mass 12 u) and oxygen (16 u) is placed in a mass
spectrometer. The ions all have the same charge and are accelerated through the same potential
difference before entering a uniform magnetic field. For each ion, the spectrometer measures the
distance between the entrance point and the place where it strikes the bottom of the region. If the
carbon ions strike the detector at a distance of 9.14 cm from the entrance point, at what distance
will oxygen ions strike the detector?
Answer: 10.6 cm
Solution: The ions travel in circular paths and land a distance of one diameter from the entrance
point. The centripetal force equation mv 2 / r = evB yields for the diameter d = 2r = 2mv / eB .
All ions have the same kinetic energy
1 mv 2
2
= eV which gives mv = 2meV . This yields for the
diameter d = 2 2mV / eB 2 ~ m , i.e. the distance is proportional to the square root of the mass.
The distance for O16 ions can then be determined by scaling the distance for C12 ions:
dO = dC mO / mC = 9.14 16 / 12 = 10.6 cm.
18. The current I in a long wire is going up as shown in the figure and
increasing in magnitude. What is the direction of the induced current in
the left loop and the right loop? List the direction of the induced current in
the left loop first. (CW = clockwise, CCW = counterclockwise)
I
Answer: CW, CCW
Solution: Using the right hand rule, we see that the B field is increasing
out of the page on the left side and increasing into the page on the right. Lenz’ law says that to
counteract those fields the induced currents must generate B fields into the page and out of the
page, respectively. Thus the currents must flow clockwise (left) and counterclockwise (right).
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PHY2054 Fall 2014
19. In the circuit shown in the figure, the switch is in position 1 for a long
time. It is then moved to position 2. How long will it take for the charge on
the capacitor to reach 50 µC after the switch is moved to its new position?
The values are C = 21 µF, R1 = 5000 Ω and R2 = 25 Ω. The battery has a
voltage of 50 V.
1 2
+
C
R2
R1
Answer: 1.60 ms
Solution: After a long time the capacitor becomes fully charged with a charge
Qmax = CV = 1050 µC . It then discharges through R2 with an RC time of 0.525 msec. Thus we
must solve 1050 ⋅ e −t /0.525 = 50 , which yields t = 1.60 msec.
20. A toroid is like a solenoid that has been bent around in a circle until its
ends meet, as shown in the figure (inner radius 4 cm, outer radius 8 cm).
What is the magnitude of the magnetic field inside a toroid of 1000 turns
carrying a current 0.5 A at a distance 6 cm from the center of the toroid?
(Hint: Use Ampere's Law applied to a closed loop in the toroid mid-plane, as
shown by a dashed line.)
Answer: 1.67 mT
Solution: We solve using Ampere’s law:
∑ i Bi ⋅ Δs = µ0ienc where Δs is a short distance along a
closed path and ienc is the enclosed current within the path. For the path we use a circular loop at
r = 6 cm where the B field is constant, obtaining B ( 2π r ) = N µ0i , where N = 1000 is the number
of turns. Solving yields B = 1.67 mT.
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