Exam 2 Solutions ```PHY2054 Spring 2012
Prof. Paul Avery
Mar. 21, 2012
Exam 2 Solutions
1. A portable CD player does not have a power rating listed, but it has a label stating that it draws
a maximum current of 159.0 mA. The player uses four 1.50 V batteries connected in series. What
is the maximum power (in milliWatts) consumed?
Solution: The power is given by P = Ei = 6 ! 0.159 = 0.954 W.
2. A galvanometer has a coil resistance of 48.0 Ω. It is to be made into a voltmeter with a fullscale deflection equal to 100.0 V. If the galvanometer deflects full scale for a current of 0.120
mA, what size resistor should be placed in series with the galvanometer?
Solution: We want the current to be 0.120 mA when the galvanometer is placed across a 100 V
source. Thus the total resistance must be 100 / 1.2 ! 10"4 = 833,000# = 0.833M# , including the
48.0 Ω resistor, which can be neglected.
3. Three resistors, R1 = 4.0 Ω, R2 = 3.0 Ω, and R3 = 2.0 Ω, are connected in series to a 9.0 V
battery. What is the power dissipated by R2?
Solution: The total current is in the series circuit is 1 A, so the power dissipated in R2 is
i 2 R2 = 12 ! 3 = 3W .
4. Three resistors, R1 = 4.0 Ω, R2 = 3.0 Ω, and R3 = 2.0 Ω, are connected in parallel to a 9.0 V
battery. What is the power dissipated by R2?
Solution: The current in R2 is 9/3 = 3 A, so the power dissipated in R2 is i 2 R2 = 32 ! 3 = 27 W .
5. In a series circuit consisting of a 12.0 V emf, a 2.00 mF capacitor, a 1.00 kΩ resistor and a
switch, what is the voltage across the capacitor at a time 2 s after the switch is closed?
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PHY2054 Spring 2012
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Solution: The voltage in the capacitor rises from 0 according to V = E 1! e!t/ RC . The RC time
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is 2 sec, thus V = 12 1! e!2/2 = 7.59 V.
6. A mass spectrometer splits a beam of 10B and 11B singly charged boron ions having the same
velocity by passing the beam through a uniform magnetic field. If the radius of the path for the
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B ions is 12.0 cm, what will be the radius for the 11B ions?
Solution: The radius is found by equating the centripetal force to the magnetic force,
mv 2 / r = evB , so r = mv / eB . Thus the radius is proportional to mass and the radius of the 11B
ion is 1.1 times the radius of the 10B ion or 13.2 cm.
7. Two long straight wires each carry a current of 12 A, as shown in
the figure. One wire lies on the x-axis with its current in the positive
x-direction. The other wire lies on the y-axis with its current in the
negative y-direction. In the x-y plane, which quadrants have points
where the magnetic field is zero?
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x
Solution: Using the right hand rule, you can see that the fields from
the wires are opposite in quadrants II and IV, so these are the only quadrants where the fields can
cancel to give 0.
8. 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 equal sides has a magnitude of:
Solution: The magnetic force is F = iLBcos 45° = 2.0 ! 0.15 ! 0.7 ! 0.707 = 0.15 N.
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PHY2054 Spring 2012
9. Refer to the previous problem. What is the magnitude of the torque on this loop due to the
magnetic field?
Solution: The torque can be found from ! = µ Bsin " = iABsin " . Using the specified values, we
obtain ! = 2 "
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" 0.152 " 0.7 " 1 = 0.01575 . You can also find the torque using the forces on the
two wires in the previous problem and the moment arms.
10. A cyclotron with a magnetic field of 1.2 T is used to accelerate protons. How does the time
for the protons to complete one orbit change when the kinetic energy increases so that the radius
of their orbit changes from 10 cm to 22 cm?
Solution: The radius of the orbit is r = mv / eB and the time to complete one orbit is
T = 2! r / v = 2! m / eB , which is independent of radius.
11. The current i in a long wire is going up as shown in the figure, but
decreasing 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.)
Solution: By the right hand rule for finding the direction of B field around a straight wire, we see
that the B field is out of the page for the left loop and into the page for the right loop. Since the
current in the wire is decreasing, by Lenz’ law the currents in the loops must reinforce the
direction of the existing fields to maintain them. Thus the currents must be counterclockwise in
the left loop and clockwise in the right loop.
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PHY2054 Spring 2012
12. In the figure, a conducting metal bar of mass 0.5 kg
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sliding along conducting rails is connected to a battery of
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emf 12 V and resistor R = 200 Ω to form a circuit. The rails X+ X
are 30 cm apart and the entire arrangement is embedded in a X X
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uniform magnetic field 0.25 T directed into the page, as
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shown. Current flowing in the circuit exerts a force on the
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metal bar, but the induced emf exerts a force in the opposite
direction. What is the maximum velocity v the bar can attain?
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Solution: The force acting on the metal bar is F = iLB = ELB / R . The bar will accelerate until
the induced emf Eind = vLB is equal to the battery emf, when the current and thus the force
becomes zero. Thus the maximum velocity is v = E / LB = 12 / ( 0.30 ! 0.25) = 160 m/s.
13. A long straight wire and a rectangular conducting loop lie in
the same plane as shown. If the current in the long wire is
decreasing, what is the direction of the net force exerted on the
loop?
Solution: The changing B field causes a current to flow in the loop. By the right hand rule for
currents, the direction of the B field is into the loop. Since the current is decreasing, Lenz’ law
says that the current must flow in a direction to reinforce the B field. Thus the current flows
counterclockwise. The magnetic forces acting on the near and far loop segments are in the
opposite direction, but the net force is not zero because the segment closest to the wire is in a
larger B field. The near segment thus determines the overall direction of force, which is upwards,
according to the right hand rule for magnetic forces.
14. An 8.0 mH inductor and a 2.5 Ω resistor are wired in series to a 23.6 V ideal battery. A
switch in the circuit is closed at time 0, at which time the current is 0. At a time 5.4 msec after
the switch is thrown the potential difference across the inductor is:
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Solution: The current in the LR circuit is V = E 1! e!tR/ L = 23.6 1! e!0.0054/0.02 = 4.4 V.
15. Refer to the previous problem. How much energy is stored in the inductor after the switch
has been closed a long time?
Solution: After a long time, the current is not changing and is i = E / R = 23.6 / 2.5 = 9.44 A. The
energy in the inductor is U = 12 Li 2 = 0.5 ! 0.008 ! 9.442 = 0.36 J.
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PHY2054 Spring 2012
16. A current of 5.0 A passes through a 1.5 H inductor. The emf in the circuit is then adjusted so
that the current changes steadily until it reaches 5.0 A in the opposite direction. During the
process, an average emf of 100 volts is measured across the inductor. How much time was
required for the current-reversal?
Solution: The emf induced in the inductor is Eind = L!i / !t = 100 V, where the average rate of
change of the current !i / !t is !i / !t = (5 + 5) / !t = 10 / !t . The time !t is the only unknown
and is easily found to be 150 ms.
17. An RL circuit with resistance 100 Ω is connected in series to a battery with emf 12 V. When
the circuit has been connected for 6.0 ms the current is measured to be 40 mA. What is the
inductance L?
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Solution: In an RL circuit, the current grows as i = imax 1! e!tR/ L , where imax = E / R = 0.12 A.
Thus e!tR/ L = 23 , which yields L = 1.480 H.
18. A solenoid of length 75 cm and diameter 8.0 cm is wound with 25 turns per cm of wire. The
current in the wire is decreasing at a rate of 150 A/s. What is the induced emf in the entire
solenoid?
Solution: The induced emf in the coil is Eind = N !" / !t , where N is the number of turns in the
solenoid and ! = " r 2 B . But the B field is given by B = µ0 ni , where n = 2500 is the number of
turns per m and i is the current. Here N = 25 ! 75 = 1875 . Putting it all together yields
Eind = N µ0! r 2 n
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"i
= 1875 # 12.56 # 10\$7 # 3.14 # 0.042 # 2500 # 150 = 4.43 V
"t
19. A hair drier has a power rating of 1500 W when plugged into a standard 120 V AC receptacle.
What is the maximum instantaneous current the hair drier must withstand?