PHYSICS
DEPARTMENT
PHY 2054
2nd Exam (SF 18–21) SOLUTIONS 6
July 2010
Each solution
is pasted in, following the list
of possible answers. Answer (1) is
the correct one.
As you can
see, the diagrams didn't make it through the file translations; some Greek
letters and sub/superscripts may
suffer similar casualties — If you don't have your copy of the exam
handy, you may have to use your
imagination!
1. Immediately after switch S in the
circuit shown is closed, the current through the battery shown is:
(1)
V_{o}/(R_{1} + R_{2}) (2) V_{o}/R_{1} (3) V_{o}/R_{2
} (4) 0 (5) V_{o}/(R_{1} + R_{2})/(R_{1}R_{2})
1.
Initially, the back emf of the inductor L blocks any current through L.
So I = V/(R1 + R2)
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 (5)
V
2.
Circuit (IV) has the
lowest potential (one battery; the parallel battery doesn't raise the voltage)
across the highest resistance. So
these bulbs are dimmest.
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
3. t = RC = (3 + 6)W (6 mF) = 54
ms . Thus t/100 = 540 ns, So i = E/R (1  e^(100) = 9v/9V [1  0] = 1A
4. The Þgure 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)
(2)
toward the magnet
(3)
away from the magnet
(4)
downwards (into the page)
(5)
along its length
4. By the
right hand rule [RHR], the force on the wire is upward.
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
Þeld B parallel to the side MK as indicated. The loop will tend to rotate:
(1)
about PQ with KL coming out of the page
(2)
about PQ with KL going into the page
(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
5. Again, by the RHR (curled fingers) the magnetic
moment of the loop is down into the page.
The torque is in a direction
so as to bring the magnetic moment into alignment with the external field
B.
So the loop rotates about
axis PQ with the top edge KL coming up out of 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 Þeld at the third vertex (P)
has magnitude (in microTeslas):
(1) 17 (2)
10 (3)
20 (4) 50 (5)
8.7
6. Each
current produces B = m_{0} I/(2 pr).
Here, r = 4 cm. The field
lines are circles and at P, the fields are *tangent* to the circles and
perpendicular to the sides of the
triangle. Thus the resultant B
is the vector sum of the fields of
the two currents :
B = 2 m_{0}
I/(2 pr) cos 30¡ = 2 (4p x10^7) 2A/(2p x 4 x 10^2) (0.866) = 1.73 x 10^5
= 17mT
7. An 8.0mH inductor and a 2.0Ω resistor are wired in series to a 20V
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 current is not given
7. After a
long time (many time constants) the inductor current is just the current
through the resistor = E/R . The
battery's potential difference appears entirely across R . So 0V, 20V.
8. An LC circuit has an oscillation
frequency of 10^{5 }Hz. If C=0.1µF, then L must be approximately:
(1)
25 µH (2) 10 mH (3)
1 mH (4) 2.5 µH (5) 1 pH
8.
At resonance , f = 1/ [2p Sqrt(LC)] . So L = 1/[(2p f)^2 C] = 1/ (39500) = 2.5 e(5) = 25m H
9. The primary of a 1:3 stepup
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
9. V1/V2
= I1/I2 = P1/P2 = N1/N2 = 1/3
. So I1 = 3I2 ==> I2 = I1/3 .
Secondary power P2 = (I2)^2 R = (I1/3)^2 R =
(I1^2)R/9 = P2/9
10. 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
10. The Bfield of the long wire is a series of circles
around the wire, going down into the page and strongest next to the wire. Thus the flux through the loop is
increasing as the wire moves toward it.
By Lenz's Law, the induced current in the loop must produce B *opposing* the change, so loop current is ccw.
11. The circuit shown is immersed in a
uniform magnetic Þeld 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
11.
There are two currents: (a) battery
current = E/R = 4V/10W = 0.4 V, ccw '
(b)
induced current = (induced emf) / R
; emf = (area) x
dB/dt = (0.12m)^2 x 150 = 2.16 V .
Then I = 2.16/10 = 0.216A. B is decreasing into the
page so I must produce B in same direction cw.
So
the two currents oppose each other: I = 0.4  0.216 = 0.18A
12. When the switch S in the circuit
shown is closed, the time constant for the growth of current in R2 is:
(1)
L/R_{2} (2) L/R_{1}
(3)
L/(R_{1} +R_{2}) (4) L(R_{1} +R_{2})/(R_{1}R_{2})
(5) (L/R_{1} + L/R_{2})/2
12. The current
through L is determined by R2, so TC = L/R2.
13. A large jetliner with a wingspan of
40 m ßies horizontally and due north at a speed of 300 m/s in a region where
the magnetic Þeld of the earth is 60µT directed 50_{◦
}below the horizontal. What is the magnitude of the induced emf
between the ends of the wing, in millivolts?
(1)
550 (2) 250 (3) 350
(4)
750 (5)
none of these
13. For
this example, E = Bwv, where w =
wingspan and B is the perpendicular B field component.
Thus E = 60 e6 T x 40 m (sin 50¡ ) x 300
m/s = 0.55 e6 = 550 e3 = 550mV
14. A ßat coil of wire consisting of 20
turns, each with an area of 50 cm_{2},
is positioned perpendicularly to a uniform magnetic Þeld 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, in milliamperes?
(1)
500
(2) 70 (3)
140 (4)
800 (5)
none of these
14. E = N dF /dt = N
(area) dB/dt = 20 x 50e4 m^2 x 2T/s = 0.2V
I =
0.2/0.4 = 0.5A = 500 mA
15. A bar magnet is falling through a
loop of wire with constant velocity. The
south pole enters Þrst. As the
magnet leaves the
loop, the induced current in the loop (as viewed from above):
(1)
is counterclockwise (2) is
clockwise (3) is zero (4) is along the length of the magnet
(5) none of these
15. As the
magnet leaves the loop, we have an Npole (B upward) moving *away* from the loop
and thus B is decreasing .
Thus
by Lenz's Law, Induced I must produce B in same direction (up) so by RHR I is CCW.
16. An incredible amount of electrical
energy passes down a funnel of a large tornado every second. Measurements taken
near the earthÕs surface in Oklahoma at a distance of 9.00 km from a large
tornado showed an almost constant horizontal magnetic Þeld of 1.50 x_{
}10^{8}T associated with the tornado. What was the average
current going down the funnel?
(1)
675 A (2) 450 A (3)
950 A (4)
1500 A (5) none of
these
16. A
horizontal field must be due to the vertical current (like that of a long
straight wire).
Thus I = 2p rB/( 4p e7) = = (rB/2) e7 = 675 A
17. A proton is released such that its
initial velocity is from right to left across the page. The protonÕs path,
however, is deßected toward the bottom edge of the page due to the presence of
a uniform magnetic Þeld. What is the direction of this Þeld?
(1)
down, into the page
(2)
up, out of the page
(3)
from the bottom edge of the page, toward the top edge
(4)
from right to left across the page
(5)
none of these
17. by the
RHR, B must be down, into the
page
18. 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
18. The voltage across R is IR = 20V, same polarity orientation as the
battery.
Moving from A
to B, we have V_{C} + 40V + 20V
= +30V ==> V_{C}= 30V
Q = CV = 50m F x 30V = 1500 mC = 1.5 mC The minus sign means
that C's voltage (and hence charge)
is opposite in polarity
to
that of R and the battery, so left plate is positive.
19. Which device(s): resistors (R), capacitors (C),
inductors (L), dissipate no energy, on average, in alternating current
circuits?
(1) C and L (2) R, C,and L (3) R only (4)
C only (5) L only
19. Only
resistors dissipate energy in ac circuits
20. Commercial electric lanterns contain
a heavyduty battery (usually 6volt) in series with an incandescent bulb and a
switch. Here are inuse electrical data from a typical lantern:
Battery terminal voltage (switch OFF): 5.6
V
Battery
terminal voltage (switch ON): 4.0
V
Circuit
current (switch ON): 0.4
A
This battery's
internal resistance in ohms is:
(1) 4 
(2) 14 
(3) 1.4 

(4) 2.4 
(5) none of
these 
20. The
terminal voltage decreased (5.6V 
4V) = 1.6V with the circuit
closed. This
corresponds to the voltage drop across the internal resistance,
so
r = V/I = 1.6/0.4 = 4 ohms