PHYSICS DEPARTMENT
2nd Exam (SF 18–21) 11 July 2006
PHY 2054
Name (print, last first):
SOLUTIONS
Signature:
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
�0 = 8.85 × 10−12C2/(Nm2)
electron charge = −1.6 × 10−19C
V=volt
N=newton
J=joule
electron mass = 9.11 × 10−31kg
m=meter
W=watt
A = ampere
“pico” = 10−12
n = “nano” = 10−9
m = “milli” = 10−3
µ = “micro-” = 10−6
proton mass = 1.67 × 10−27 kg
c = 3 × 108 m/s
Answer (1) is the correct answer. The (brief) solution is in bold italics
like this. [NOTE: ‘RHR’ = “Right Hand Rule”]
1. Immediately after switch S in the circuit shown is closed, the current through the battery
shown is:
At t=0 , L acts like an open circuit (due to the initial back emf) so initial
current flows only through R1 and R2
(1) Vo/(R1 + R2)
(2) Vo/R1
(3) Vo/R2
(4) 0
2. In the diagrams, all light bulbs are identical and all emf devices are identical. In
which circuit (I, II, III, IV, V) will the bulbs be dimmest?
(1) IV
(2) I (3) II
(4) III
(IV) has 2 bulbs in series across a single emf source, hence the lowest current => dimmest bulbs.
3. 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)1A
(2)1/2A
(3)3/4A
(4)3/8A
(5)3/2A
i= V/R e-t/τ but t/τ = (τ/100)/ τ = 1/100 so I = V/R e-100) = V/R = 9V/(6 Ω + 3 Ω) = 1A
4. 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 (out of the page) By the right–hand rule, since B is directed away from the
(2) toward the magnet
magnet toward the (-) current, the force on the current is
(3) away from the magnet
upward.
(4) downwards (into the page)
(5) along its length
5. 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) about PQ with KL coming out of the page RHR: force on KL is upward & is
(2) about PQ with KL going into the page
downward on MN
(3) about RS with MK coming out of the page
(4) about RS with MK going into the page
(5) about an axis perpendicular to the page
6. Two long straight wires pierce the plane of the paper at vertices of an equilateral triangle as
shown at right. They each carry 2 A, out of the page. The magnetic field at the third vertex
(P) has magnitude (in T):
5
5
5
6
(1) 1.7 10− (2) 1.0 10− (3) 2.0 10− (4) 5.0 10−
×
×
×
×
At P, B from the two lower currents is the vector sum of r) from each current =>
B = 2 (0.04m] cos 30° = 1.7 E-5 T
7. Magnetic field lines inside the solenoid shown are:
(1) upward, out of the cylinder
RHR: B due to I is upward!
(2) clockwise circles as one looks down the axis from the top of the page
(3) counterclockwise circles as one looks down the axis from the top of the page
(4) downward, into the cylinder
(5) in no direction since B = 0
8. 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, respectively:
(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
After many time constants, the voltage across L is zero, hence all of the emf appears across R
5
9. An LC circuit has an oscillation frequency of 10 Hz. IfC=0.1µF, then L must be about:
(1) 25 µH (2) 10 mH (3) 1 mH (4) 2.5 µH (5) 1 pH
f = 1/ [2LC)] => L = 1/([2f2C)
10. The primary of a 3:1 step-up transformer is connected to an AC source and the secondary is connected to a resistor R. The
power dissipated by R in this situation is P. If R is connected directly to the source it will dissipate a power of:
(1) P/9 (2) P/3
(3) P
(4) 3P
(5) 9P
A 3:1 step-up=> V2 = 3V1 . So P2 = (3V1)2 / R = 9V12/R . But if R
is connected directly to V1 the power is just V12/R = P/9
11. A long straight wire is in the plane of a rectangular conducting loop. The straight wire carries
a constant current i, as shown. If the wire is moved toward the rectangle at constant speed, the
induced current in the rectangle is:
(1) counterclockwise
(2) zero
(3) clockwise
(4) clockwise in the left side and counterclockwise in the right side
(5) counterclockwise in the left side and clockwise in the right side
The B field due to I is into the page at the loop. The flux is increasing since B is
greater near the loop. The induced emf must produce B
that opposes the change, hence induced B is out of the
page => induced I is ccw [RHR]
12. The 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
The current in the circuit by itself (without the changing B) is I= (4V)/10Ω = 0.4A .
The induced emf = A(dB/dt)= (0.12m)2 (150 T/s) = 2.16V => induced I = 2.16V/10Ω = 0.216A.
Since B is decreasing into the page, the emf must induce a current that produces B into the page, hence
the induced current is cw, opposing the above current. so total I = 0.4A - 0.216A = 0.18A
13. When the switch S in the circuit shown is closed, the time constant for the growth of
current in R2 is:
(1) L/R2 (2) L/R1 (3) L/(R1 +R2) (4) L(R1 +R2)/(R1R2) (5) (L/R1 + L/R2)/2
Only R2 affects the current in L, hence τ= L/R2
14. A large jetliner with a wingspan of 40 m flies horizontally and due north at a speed of 300 m/s in a region where the
magnetic field of the earth is 60µT directed 50◦ below the horizontal. What is the magnitude of the induced emf between the
ends of the wing?
(1) 550 mV (2) 250 mV (3) 350 mV (4) 750 mV (5) none of these
emf = Blv = 60µT sin 50 (40m) (300 m/s)
2
15. A flat coil of wire consisting of 20 turns, each with an area of 50 cm , is positioned perpendicularly to a uniform magnetic
field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40Ω,
what is the magnitude of the induced current?
(1) 500 mA (2) 70 mA (3) 140 mA (4) 800 mA (5) none of these
emf = NAdB/dt = 20 (0.005m2)[4T/2s] = 0.2V. Thus i = 0.2V/0.40Ω = 0.5A = 500 mA
16. A bar magnet is falling through a loop of wire with constant velocity.
leaves the wire, the induced current in the loop (as viewed from above):
The south pole enters first. As the magnet
(1) is counterclockwise (2) is clockwise (3) is zero (4) is along the length of the magnet (5) none of these
As the N pole leaves the loop, B is upward & decreasing. By Lenz's law & RHR, induced i must produce
B in same direction of B (upward) so I is ccw
17. An incredible amount of electrical energy passes down a funnel of a large tornado every second.
Measurements taken
in Oklahoma at a distance of 9.00 km from a large tornado showed an almost constant horizontal magnetic field of 1.50
×
8
10− T associated with the tornado. What was the average current going down the funnel? (µ0 =
7
4π 10− T m/A)
×
·
(1) 675 A (2) 450 A (3) 950 A (4) 1500 A (5) none of these
B =r => I =r B/ 
18. A proton is released such that its initial velocity is from right to left across the page. The proton’s path, however, is deflected
toward the bottom edge of the page due to the presence of a uniform magnetic field. What is the direction of this field?
(1) into the page
(2) out of the page
(3) from the bottom edge of the page
(4) from right to left across the page
(5) none of these
By the RHR, downward magnetic force on a + particle moving right to left must be due to B directed into page.
19. If I =2.0 mA and the potential difference, VA VB = +30V in the
−
circuit segment shown, determine the charge and polarity of the
capacitor.
(1) 1.5 mC, left plate is positive
(2) 1.5 mC, right plate is positive
(3) 0.50 mC, left plate is positive
(4) 0.50 mC, right plate is positive
(5) none of these
The current is from A to B. Hence the loop equation (going from A to B) is (guessing the polarity of C):
-(Q/C) -40V - 10K (2ma) +30V = 0. Thus Q/C = -30V => Q= -1.5mC. The - sign means that the
polarity of C is opposite to what we guessed in the loop equation, so left side is +.
20. The ‘rms’ values of voltage and current employed in alternating-current (AC) calculations are useful for:
(1) determining the power required to operate AC electrical devices.
(2) making the subject more difficult for students.
(3) finding the frequency of the associated electromagnetic waves.
(4) calculating the subterranean structural mass of arboreal flora.
(5) none of these
Surely this one is obvious!