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Instructor(s): F.E. Dunnam
PHYSICS DEPARTMENT
PHY 2054
2nd Exam
08 July 2008
Signature:
Name (print, last first):
On my honor, I have neither given nor received unauthorized aid on this examination.
YOUR TEST NUMBER IS THE 5-DIGIT NUMBER AT THE TOP OF EACH PAGE.
(1) Code your test number on your answer sheet (use 76–80 for the 5-digit number). Code your name on your
answer sheet. DARKEN CIRCLES COMPLETELY. Code your UFID number on your answer sheet.
(2) Print your name on this sheet and sign it also.
(3) Do all scratch work anywhere on this exam that you like. Circle your answers on the test form. At the end of
the test, this exam printout is to be turned in. No credit will be given without both answer sheet and printout.
(4) Blacken the circle of your intended answer completely, using a #2 pencil or blue or black ink. Do not
make any stray marks or some answers may be counted as incorrect.
(5) The answers are rounded off. Choose the closest to exact. There is no penalty for guessing. If you believe that no
correct answer is listed, leave this item blank!!
(6) Hand in the answer sheet separately.
Useful (??) Constants:
µ0 = 4π × 10−7 Tm/A
²0 = 8.85 × 10−12 C2 /(Nm2 )
−19
electron charge = −1.6 × 10 C
electron mass = 9.11 × 10−31 kg
V=volt
N=newton
J=joule
m=meter
W=watt µ = “micro-” = 10−6
A = ampere “pico” = 10−12 n = “nano” = 10−9 m = “milli” = 10−3 proton mass = 1.67 × 10−27 kg
c = 3 × 108 m/s
1. Immediately after switch S in the circuit shown is closed, the current through the battery shown is:
(1) Vo /(R1 + R2 )
(2) Vo /R1
(3) Vo /R2
(4) 0
(5) Vo (R1 + R2 )/(R1 R2 )
SOLUTION: At the instant of circuit-closure, L’s back emf allows no current to pass through it, so I = Vo /(R1 + R2 )
2. For the above circuit, a long time after the switch is closed, the current through the inductor will be approximately
(1) Vo /R1
(2) Vo /(R1 + R2 )
(3) Vo /R2
(4) 0
(5) Vo (R1 + R2 )/(R1 R2 )
SOLUTION: After a long time, L’s back emf is zero, so V R1 = 0 and thus I = Vo /R1
3. When switch S is open, the ammeter in the circuit shown reads
2.0 A. When S is closed, the ammeter reading will:
(1) increase slightly
(2) remain the same
(3) decrease slightly
(4) double
(5) halve
SOLUTION: Closing S provides another conducting path, reducing the total R slightly, so A increases slightly
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4. In the circuit shown, the capacitor is initially uncharged and V =
9.0 volts. At time t = 0, switch S is closed. If τ denotes the time
constant, the approximate current through the 3Ω resistor when
t = τ /100 is:
(1) 1 A
(2) 1/2 A
(3) 3/4 A
(4) 3/8 A
(5) 3/2 A
SOLUTION: I = E/R[1 − e∗∗ (−t/RC)]. The time constant = 1/RC. If 1/RC = 1/100, then we have I =
9v/9Ω[1 − e∗∗ (−100)] = 1A[1 − 0] = 1A
5. An electron is moving southward in a region where the magnetic field is directed northward. The magnetic force
exerted on the electron is:
(1) zero
(2) up
(3) down
(4) east
(5) west
SOLUTION: If v is parallel to B, the force is zero!
6. A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On
the way, the electrons encounter a magnetic field directed vertically downward. The spot on the screen will therefore
be deflected:
(1)
(2)
(3)
(4)
(5)
to the right as seen from the electron source
upward
downward
to the left as seen from the electron source
not at all
SOLUTION: By the right-hand-rule, the beam is deflected to the right
7. The figure shows the motion of electrons in a wire which is near
the N pole of a magnet. The wire will be pushed:
(1) upwards
(2) toward the magnet
(3) away from the magnet
(4) downwards
(5) along its length
SOLUTION: Right-hand-rule: wire is pushed upward
8. A square loop of wire lies in the plane of the page and carries a
current I as shown. There is a uniform magnetic field B parallel to
the side MK as indicated. The loop will tend to rotate:
(1)
(2)
(3)
(4)
(5)
about
about
about
about
about
PQ with KL coming out of the page
PQ with KL going into the page
RS with MK coming out of the page
RS with MK going into the page
an axis perpendicular to the page
SOLUTION: Right-hand-rule: force on segment KL is up, on NM is down, so it rotates about PQ with KL moving
upward
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9. The magnitude of the magnetic field at point P, at the center of
the semicircle shown, is given by:
(1) µ0 i/4R
(2) µ0 i/R2
(3) µ0 i/2πR
(4) µ0 i/4πR
(5) µ0 i/2R
SOLUTION: This is half of a circular current loop, so I = 12 [µ0 i/2R] = µ0 i/4R
10. Two long straight wires pierce the plane of the paper at vertices
of an equilateral triangle as shown below. They each carry 2 A,
out of the paper. The magnetic field at the third vertex (P) has
magnitude (in T):
(1) 1.7 × 10−5
(2) 1.0 × 10−5
(3) 2.0 × 10−5
(4) 5.0 × 10−6
(5) 8.7 × 10−6
SOLUTION: Each of the currents produces a field at P equal to B = µ0 i/2πR, but each of these is perpendicular
to its side of the triangle at P. The net field is just the sum of the horizontal components: B = 2µ0 (2A)/2π(4 ×
10−2 m) cos 30◦ = 1.73 × 10−5 T
11. Magnetic field lines inside the solenoid shown are:
(1)
(2)
(3)
(4)
(5)
toward the top of the page
clockwise circles as one looks down the axis from the top of the page
counterclockwise circles as one looks down the axis from the top of the page
toward the bottom of the page
in no direction since B = 0
SOLUTION: RHR again, B is upward
12. Four long straight wires carry equal currents into the page as
shown. The magnetic force exerted on wire F is:
(1) east
(2) north
(3) south
(4) west
(5) zero
SOLUTION: The forces are all attractive and vertical components cancel. The net force thus is to the right, or
eastward.
13. An 8.0-mH inductor and a 2.0-Ω resistor are wired in series to a 20-V ideal battery. A switch in the circuit is closed
at time 0, at which time the current is 0. A long time after the switch is thrown the potential differences across the
inductor and resistor are:
(1) 0, 20 V (2) 20 V, 0 (3) 10 V, 10 V (4) 16 V, 4 V (5) unknown since the rate of change of the current is not given
SOLUTION: After a long time, L’s current is max, and its resistance is negligible. The only ∆V thus is across R,
so ∆VL = 0; ∆VR = 20V
14. An LC circuit has an oscillation frequency of 105 Hz. If C = 0.1µF, then L must be about:
(1) 25 µH
SOLUTION: fR =
(2) 10 mH
(3) 1 mH
(4) 2.5 µH
1
1
1
√
L=
× 10−7 F = 25 × 10−6 H
≥ LC =
2
2
2
4π f
4π × (105 )2
2π LC
(5) 1 pH
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15. A long straight wire lies in the plane of a rectangular conducting
loop as shown. If the current i in the wire is decreasing, the current
in the loop will be
(1)
(2)
(3)
(4)
(5)
clockwise
zero
counterclockwise
clockwise in the left side and counterclockwise in the right side
counterclockwise in the left side and clockwise in the right side
SOLUTION: B due to current is down into page. By Lenz’s law, since it’s decreasing, induced emf in loop must
reinforce B, so by RHR loop current is clockwise
16. The entire circuit shown is immersed in a uniform magnetic field
that is into the page and is decreasing in magnitude at the rate 150
T/s. The current in the circuit (in amperes) is:
(1) 0.18
(2) 0.22
(3) 0.40
(4) 0.62
(5) none of these
SOLUTION: Without the flux change I = 4V /10Ω = 0.4A, CCW Induced V = dΦ/dt = AdB/dt = (0.12m)2 ×
150T /s = 2.16V ; the induced current is then 2.16V /10Ω = 0.216A. B is decreasing so to reinforce it [Lenz’s law] this
induced current must be clockwise. The net current is thus 0.4A − 0.216A = 0.18A
17. A rod with resistance R lies across frictionless conducting rails
in a uniform magnetic field B, as shown. Assume the rails have
negligible resistance. The force that must be applied by a person
to pull the rod to the right at constant speed v is:
(1) B2 L2 v/R
(2) 0
(3) BLv
(4) BLv/R
(5) B2 Lxv/R
SOLUTION: The induced voltage = BLv. At constant speed, the ohmic heating V2 /R = work done/t = Fx/t, =
Fv, so F = B2 L2 v / R
18. In the circuit segment shown, the current I = 2.0 mA, flowing
from A to B, and the potential difference, VA − VB is +30V. At
this instant, the charge and polarity of the capacitor C will be:
(1)
(2)
(3)
(4)
(5)
1.5 mC, left plate is positive
1.5 mC, right plate is positive
0.50 mC, left plate is positive
0.50 mC, right plate is positive
none of these
SOLUTION: VA − VB = 30V . Applying the loop rule: 30V + Q/C − 40V − 10kΩ(2mA) = 0 Q/C = 30V ≥ Q =
30V (50 × 10−6 F ) = 1.5mC. In traversing the loop we took Q/C as + , =¿ outgoing (left) side is +.
19. In an ideal 1:8 step-down transformer the primary power is 10 kW and the secondary current is 25 A. The primary
voltage must therefore be
(1) 3200
(2) 25,600
(3) 400
(4) 50
(5) 6.25
SOLUTION: Pin = Pout ≥ V1 I1 = V2 I2 and for 8:1 step down, V1 = 8V2 So 10 kW = 25 V2 ≥ V2 = 10, 000 W/25V
= 400 V V1 = 8 (400V) = 3200 V
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20. Electric utility companies provide alternating current (ac) instead of direct current (dc), primarily because
(1)
(2)
(3)
(4)
(5)
ac voltages may be conveniently transformed
ac is less lethal
ac minimizes magnetic effects
dc-powered incandescent bulbs are dimmer
the batteries required for large dc are prohibitively expensive
SOLUTION: ac can be conveniently transformed (up or down), making transport of high power more efficient: high
voltage ≥ lower current ≥ less ohmic loss in the lines.