1
Final Exam, vers. 0001 - Physics 1120 - Fall, 2006
Solutions:
1. An electromagnetic plane wave with intensity I propagates through space and is incident on
an absorbing panel as shown in the figure below. The surface of the panel is parallel to the y-z
plane, is centered at the point xo, has height h and width w. The plate absorbs all of the energy
that is incident upon it.
What is the power absorbed by the panel?
A)
B)
C)
D)
E)
Iwh
Iwh/5
I w h / 25
4 I w h / 25
Zero
Intensity is just "power/area", so power = I*area.
The plane wave is uniform, same everywhere, so
it's just I*total area of the plate. (This is like a
solar panel! If you double the area of the panel,
you double the energy you receive)
2. When a violin is played, the bow causes the
string to vibrate. The vibrating string causes the
body to vibrate and thus sound emanates from the
body of the instrument. Consider the following
statements:
I. Light waves are like the waves on the string because both are transverse waves.
II. Light waves are like the waves on the string because light waves can only propagate in one
dimension (along a single line).
III. Light waves are like the sound waves because light waves propagate in three dimensions.
Which statement(s) are correct?
A)
B)
C)
D)
E)
I only
II only
III only
I and II
I and III
Light waves ARE transverse, as are waves on a string. This is one key thing they have in common
Light waves DO propagate in 3-D, must like sound waves. This is another key commonality.
(II is no good because it's surely not remotely correct to say light waves "can only" propagate in
1-D) Even a laser beam spreads out some!
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3. Complete the following analogy relation:
Sound is to Air Pressure as Light is to ___
A) Vacuum
B) Electrons
C) Vectors
D) Electromagnetic field
E) Color
Sound is a wave, the quantity which is varying is the air pressure.
Light is a wave, the quantity which is varying is the EM fields (both E and B)
4. Which schematic diagram best represents the realistic circuit shown below
This circuit has two bulbs in SERIES with one another (at the top), and a third
bulb which is in PARALLEL to the two of those (i.e. directly connected across the
battery) that's A below, none of the others is the same!
A)
B)
D
)
C)
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5. What happens to the potential difference between points 1 and 2 if bulb A is removed?
A
B
1
A)
B)
C)
D)
2
It increases
It decreases
It stays the same
The answer depends on whether bulb A has a greater or smaller resistance than B
Point 1 is connected via ideal wire to one side of a battery, point 2 is connected to
the other. The voltage drop from 1 to 2 must be exactly V(batt!) Doesn't matter
what A is doing there...
6. All the batteries and bulbs in this question are identical. Compare the brightness of bulbs A
and B in circuit 1 with the brightness of bulb C in circuit 2.
A
B
C
Circuit 1
Circuit 2
What is the correct ranking of the brightness of the bulbs?
A)
B)
C)
D)
E)
A=B>C
A=B<C
A=C<B
A>B>C
A=C>B
There is an ideal wire in the middle of circuit 1, that "short circuits" bulb B. B is dark, it doesn't
glow at all. So A is just "across" the battery, just like C. A and C are equally bright, B is off.
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The next two questions refer to this circuit.
Two identical bulbs and a capacitor are connected to an ideal battery of voltage V as shown.
Initially, the switch is open and the capacitor is uncharged.
A
V
1
B
C
2
7. The switch is closed. Which one of the following best describes the behavior of the brightness
of bulb B as a function of time from the point immediately after the switch is closed
A)
B)
C)
D)
E)
It starts off completely dark and stays completely dark
It starts off bright and gets dimmer until it goes completely dark
It starts off bright and stays bright
It starts off completely dark and gets brighter
If starts off bright and gets a little dimmer, but never goes completely dark
This is JUST like an old exam question!! At t<0, C is uncharged, so
(since Q = C Delta V) there is no voltage drop across the capacitor. (V1=V2).
At t=0+, the charge on the capacitor can't change instantly, it takes time.
So Delta V is STILL zero, for an instant, that means V1=V2 still, and so bulb B
has no voltage drop across it. It is dark. All the current is going through A and
right into the capacitor (which acts like a "short" at first, another way to think
about what's happening) As time goes by, though, charge builds up, so V1>V2, and
current starts to also flow through B. It gets brighter.
8. After the switch has been left closed for a very long time, what is the magnitude of the voltage
difference between points 1 and 2?
A) 2V
B) 0
C) V
D) V/2
After a long time, the capacitor is charged and no current flows through it. So all
current goes through the A-B series. In the middle of a series circuit with two
identical bulbs, the voltage is half (half the drop across the first, half the drop
across the oterh).
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9. A circular current loop encloses a long straight wire. The long
straight wire carries a current out of the page which is increasing
with time. The induced current in the wire loop
A) is in the clockwise direction
B) is in the counterclockwise direction
wire loop
Straight wire
current I
out of page
C) is zero
The wire produces a B field which is "loops" around itself. Since it is very long, these loops are
purely in the plane of the page. That means the B field created by the wire is IN THE PLANE OF
THE PAGE, which means there is NO FLUX "cutting" through that loop! As the current
changes, the field is increasing, but the flux is still zero, and stays zero.
No change in flux means no current induced.
The loop is in the wrong orientation to get any current flowing in it!
10. A coil of wire with 100 turns and area 0.01 m2 is connected
to the two ends of a 3 Ohm resistor. The wire is covered in
varnish so that the turns do not short-circuit. The coil surrounds a
solenoid which has 30 turns per cm, a cross-sectional area of
0.005 m2, and carries a current which is increasing at a rate of
100 A/s. What power is being dissipated by the resistor?
[Choose the closest answer]
A) 4.7x10-4 W
C) 1.2 W
E) 0.012 W
B) 1.2x10-4 W
D) 0.047 W
B
increasing
Long solenoid of
circular cross
section going into
and out of the page
resistor
(3 Ohm)
Coil of
wire
(100
turns)
Power is V^2/R, so we need to find the voltage (or EMF). That has to be Faraday's
law!
EMF = - N d(phi)/dt, where phi = magnetic flux = B*area.
Here, B = B(solenoid) = mu0 * n * I.
Here, Area = "area where there is a B field" = "area of solenoid" (NOT the area
of the coil - there is no B field in much of that area!)
So EMF = -100 turns * d/dt ( mu0 * n * I * small area)
= -100 * 4pi*10^-7 * (30 turns/cm* 100cm/meter)*(dI/dt) * .005 m2
= .188 Volts. (I used dI/dt = 100 A/s)
Then, Power = .188V^2/3 Ohm = .012 W
Wow, we needed a calculator! (Wanted to make sure everyone could keep track of
powers of 10 and stuff :-)
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11. A solenoid has a decreasing current causing a decreasing uniform B-field in its interior. An
end-on view of the solenoid is shown to the right. What is the direction of the electric field at
the point x inside the solenoid and directly below the solenoid’s
solenoid
central axis as shown in the diagram?
(A) left ←
(B) right →
(C) up ↑
(D) down ↓
(E) zero
B (into page)
This is Lenz' law. The field is into the page but DECREASING,
point x
which means the CHANGE in that field is out of the page. To
fight that change we induce EMF's in the clockwise direction
(use your right hand, convince yourself!) At point x,
clockwise means "left"
12. A small elliptical shaped wire loop is carrying a
current I which is increasing in time. Above it and
below it are two small circular loops, labeled 1 and
I, increasing
2, as shown. All three loops are in the plane of the
page. The currents induced in loop 1 and loop 2 are
loop 1
(A) both clockwise
loop 2
(B) both counterclockwise
(C) both zero
(D) clockwise
in loop 1 and counterclockwise
(E) counterclockwise
in loop 1 and clockwise
in loop 2
in loop 2
This is just Lenz' law. The only trick is to first think hard about the field PRODUCED by the oval
current, at the two loop locations!
What field does that oval current make? Using the RHR, it makes a field which comes OUT of
the page thoughout the middle of the oval, and INTO the page everywhere else. (If you're not
convinced that this works for an oval, think of the field made by the upper part of the oval wire
alone. It goes into the page at the top of the picture, and out of the page below. The other wire
does the opposite. But if you're up at loop 1, the lower part of the oval is farther from you, so the
upper part of the oval "wins": so as I said, net B field up there is into the page. Similarly, down
where loop 2 is, the "lower part of the oval" wins, and the net field is, again into the page.
So B is "into the page" and increasing at both loops. Lenz' law says "fight that change", that
means a counterclockwise current (in both!)
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13. The figure to the right shows two charges, -2Q, and +Q a distance 2d apart.
-2Q
What would be the magnitude of the force on a charge q if it were placed at the
point P which is a distance d from the +Q charge as shown
(A)
4 kQq
3 d2
(B)
17 kQq
9 d2
(C)
2 kQq
3 d2
(D)
7 kQq
9 d2
2d
+Q
+d
(E) zero
P
This is just Coulomb's law, F = kqQ/r^2, added up (as a vector!) for all charges.
Here, we get kqQ/d^2 DOWN and kq(2Q)/(2d+d)^2 UP, which comes out to
ksq/d^2*(1-2*1/9), which is D.
14. The figure to the right shows seven point charges, all of
equal magnitude. Five are positive charges and two are
negative charges. Determine the direction of the total force
on the charge at the origin due to all other charges shown.
Into which quadrant does the total force vector point?
y
D
A
(A) Quadrant A, that is, upwards but slanted to the right
x
(B) Quadrant B, that is, downwards but slanted to the right
(C) Quadrant C, that is, downwards but slanted to the left
(D) Quadrant D, that is, upwards but slanted to the left
C
(E) There is not enough information given to decide for sure
The two minus charges pull you into quadrant A. The two + charges left and down
ALSO push you into quadrant A. The + below you and the + above you "fight", but
the lower one wins, so there's an additional "upward" push. Adding many vectors
which are either up, right, or up and right must yield a sum which has both up and
right components, which means quadrant A!
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15. Which one of the follow statements is correct (only one is correct)?
(A) The net magnetic flux through a closed surface may be non-zero
No, this is false, if the magnetic flux through a closed surface was zero, there would have to be a
magnetic "charge" (or monopole) inside... and to the best of our knowledge, after 150 years
looking there are no such beasts in the universe! (If you discover that the answer to this question
is ever true, you will win the Nobel Prize!!)
(B) B-field lines may cross
No, this wouldn't make any sense. What would it MEAN if field lines cross? What would the field
B at the crossing point??
(C) The magnetic force on a charged particle may change its kinetic energy
No, the magnetic force is always perpendicular to v (that's the right hand rule), and that means it
never does any work, which means no change in KE.
(D) A test charge released from rest will always initially move along a B-field line.
No, a test charge at rest is not influenced by B fields, you need a velocity!
(E) Where the B-field lines are most dense, the magnitude of the magnetic field is largest
Yup, that's right - it's how we interpret field line diagrams! (Same for E-field diagrams too)
The next two questions refer to this situation. Four point charges, of
equal charge magnitude Q are located at the corner of a square. Point P
is at the center of the square. Two charges are positive, and two
negative as shown.
16. What is the magnitude of the total electric field at a point P?
(A)
8 kQ
2
2 L
kQ
L
(E) none of these
(C) 4
(B)
4 kQ
2
2 L
(D) 8
Q
-Q
P
Q
-Q
L
kQ
L2
Each charge makes a field at P which is "diagonal", and ALL point either "up and right" or
"down and right". So the y components cancel, and we're just adding the x components. But all
four of those are the same! So figure out 1, and multiply by 4!
E_x(from one Q) = k Q/(distance^2)*cos(angle with x axis)
Here, distance is HALF the diagonal of the square, i.e. (1/2) L*Sqrt[2].
Distance squared is thus L^2/2
Here, angle is 45 degrees, and cos(45) = 1/sqrt[2]
So I get 4E_x =[ 4*k Q /(L^2/2) ] * (1/sqrt[2])
That's 8 kQ/Sqrt[2] * 1/L^2.
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17. What is the direction of the force on an electron placed at point P due to the four point
charges?
(A) up, along +y direction
A
(B) right, along +x direction
(C) down, along –y.
D
B
(D) left, along –x
C
(E) some other direction, or the net force is zero.
Already worked this out (above), the field is to the right. But an
electron has negative charge, so the FORCE on it is to the left (attracted to the +'s, repelled by
the -'s, makes sense!)
18. A charged particle with mass m and charge q is at rest at the
origin. A uniform electric field is present of magnitude E and with a
direction which makes an angle ! with the positive y-axis, as
shown. What is the x-component of particle’s acceleration?
(A) q E m sin(! )
q E sin(! )
m
(E) none of these
(C)
y
!
(B) q F m cos(! )
(D)
r
E
q E cos(! )
m
m, q
x
F=ma is Newton's law.
F = qE is from our basic definition of electric field.
So a = qE/m. Then, the x-component will be (look at the picture!)
a_x = (qE/m) * sin(theta)
!
19. A ray of light passes from air through the corner of a glass block so
that its angle of entry and angle of exit, ! , are exactly the same as
shown in the diagram. Assuming the refractive index of air to be 1, and
the refractive index of glass to be n, what is the angle ! ?
& n #
& n #
(A) sin '1 $
(B) sin !1 2 n
(C) cos '1 $
!
!
% 2"
% 2"
!
( )
( )
(D) cos !1 2 n
(E) none of these
The figure shows the SAME angle coming in and going out. So by Snell's law, whatever that
"inner angle" is, it must also be the SAME at the entry and exit points. So the little triangle up
there has two equal angles, they must both be 45 degrees! Then, just use
n(air)*sin(theta) = n(glass)*sin(45 degrees),
which means theta = sin^-1(n*sin(45 degrees))
= sin^-1(n/sqrt[2])
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20. An object is located to the left and
above the optic axis of a thin converging
lens. The focal points are labeled f. Where
will the image be located?
(A) Location A
(B) Location B
(C) Location C
(D) At the right focal point
(E) There will not be a image
Object
C
f
f
A
B
D
Draw the principle rays, and see where they
meet! It's point B.
21. A step-down transformer is attached to an AC voltage source and a resistor as shown. There
are 12 turns on the primary coil and 4 turns on the secondary coil. Assuming the transformer is
ideal (this means that no power is lost), how does the rms current through the resistor IR compare
with the input current Iin?
(A) IR= 4Iin
(B) IR= (1/4)Iin
(C) IR= 3Iin
(D) IR= (1/3)Iin
Iin
IR
(E) IR= 0
Step down means the voltage DROPS by the ratio of
turns, here 12:4 = 3:1
Conservation of power (IV) tells me that current will then go UP by 3:1
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iron core
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22. Two ideal polaroid filters, labeled 1, and 2, are placed between a source of unpolarized light
and an observer. As shown, the pass axes of the two filters are at angles of 0o and 90o relative to
the vertical. The filter 2 is then removed. What does the observer see?
light source
(unpolarized)
polarizer 1
observer
2
(A) No light when filter 2 is present and some light when filter 2 is removed.
(B) Some light when filter 2 is present and no light when filter 2 is removed.
(C) No light when filter 2 is present and still no light when filter 2 is removed.
(D) Some light when filter 2 is present and more light when filter 2 is removed.
(E) Some light when filter 2 is present and the same amount of light when filter 2 is removed.
The first polarizer cuts the intensity in half, and leaves the light polarized. Then the light hits #2,
which is "crossed", so NOTHING comes through. When you take away #2, you have that "half
intensity light".
23. A 1200W heater is attached to 120VAC circuit. What of the following is closest to the peak
current through the heater?
A) 20A
B) 14A
C) 10A
D) 7A
E) 0A
Average power = I(max)*V(max)/2 = I(rms)*V(rms)
Here, 120 VAC means "RMS", that's the convention we've used.
So I(rms) = 1200 W / 120 V = 10 Amps. The question asks for peak, which is
Sqrt[2] bigger, 10*Sqrt[2] = 14 A.
24. Consider a solid metal conducting sphere of radius R with a charge Q . If the voltage is
defined to be zero an infinite distance from the sphere, what is the voltage at a point half way
between the center of the sphere and the surface?
2kQ
8kQ
kQ
kQ
4kQ
(A)
(B)
(C)
(D)
(E)
R
R
R
2R
R
OUTSIDE of the sphere, V = kQ/R, that's our familiar old formula. (Once you're outside, there's
no way to tell whether there's a sphere or a point "inside" you)
But E=0 INSIDE of conductors (in equilibrium), and that means V=constant everywhere INSIDE
the conductor. So there's no change, it's still kQ/R!
Note that if it was an INSULATING sphere, AND if we told you the charge was uniformly
distributed throught the whole volume, THEN you could do (like we did in a CAPA problem) and
figure out E field and thus voltage everywhere inside. (And then a factor of 2 might appear!) But
that's a different problem!
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The next two questions refer to this situation. Two charges, each of equal charge q , are a
distance 2a apart as shown.
25. What is the electric potential at point P, half way between the charges?
(A) 0
kq
a2
2k q
(E) 2
a
(C)
kq
a
2k q
(D)
a
(B)
point P
q
a
a
Voltage is just the sum of kq/r, here it's 2kq/a. (Note that voltages add like
numbers, not vectors, they do not "cancel")
26. If the charges are initially well separated, and then assembled in the configuration shown by
an external agent, how much work did the external agent have to do?
k q2
k q2
2k q 2
k q2
kq
(A)
(B)
(C)
(D)
(E)
2
2
2a
a
2a
4a
4a
It takes ZERO work to bring the first charge (say, the left one) to its spot, if the other one is still
at infinity.
Then, to bring the second charge in, extra Work = change in potential energy =
q delta V = q*(kq/2a) (The factor of 2 is because the distance BETWEEN them is 2a)
The next three questions refer to this
R1 = 10!
situation. A circuit consists of a battery, two
resistors, an inductor and a switch as shown.
Initially the switch is open and has been open V = 10V
for a long time
L = 10H
R2 = 10!
27. The switch is closed. Immediately after
the switch is closed, what is the magnitude of
the current that flows through the battery?
A) 0 A
B) 1 A
C) 0.5 A
D) 2 A
Immediately BEFORE, there was NO current through L. That can't suddenly change. So
immediately AFTER, no current goes through L, it must all be going around the big outer loop..
That's a series circuit of two 10 Ohm resistors, total R=20 Ohm, so current = 10 V/20 Ohm =
0.5 A
28. After a long time, what is the magnitude of the current through the inductor?
1
1
A) 1 A
B) 0 A
C)
A
D) 100 A
E)
A
10
100
After a LONG time, the inductor looks like an ideal wire, it is "shorting out" R2. So
you now have ONE resistor, R1, and current = 10 V / 10 Ohm = 1 Amp.
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29. After being left closed for a very long time the switch is opened again. Immediately after the
switch is opened, what is the magnitude of the voltage difference across resistor R2 ?
A) 0 V
B) 5V
C) 10V
D) 20V
E) infinity
Just BEFORE we make this change, the current (we just found) is 1 Amp. Just AFTER, the
current stays the same (current can't suddenly change in an inductor) so you have 1 Amp still
flowing. Where can it go? The switch is open, so it HAS to go the other way, through R2=10
Ohm. 1 Amp flowing through 10 Ohms must have a delta V = IR = 1*10 = 10 V
30. An electron is released at a point A near one plate of a parallel plate
capacitor and moves to a point B near the other plate as shown. If the
voltage V of the battery is 10 V, what is the change in electrostatic potential
energy "U = U B ! U A of the electron?
(A) 10 J
(B) -10 J
(D) ! 1.6 " 10 !18 J
(E) zero
A
B
(C) 1.6 " 10 !18 J
V
U = qV, so here we have Delta U = qDelta V
q is the charge of the electron (-e), and Delta V = Vb-Va = +10 V
(Look at the way the battery is hooked up, the higher, + side is on the "B" side"
So Delta U = (-e)*(10V-0) = -1.6E-19*10 = -1.6E-18 J
Does it make sense? Electrons would be attracted to the + plate, they would spontaneously go
fro A to B. Objects like to go to lower potential energy (like dropping a rock!) so it totally makes
sense that the answer should be NEGATIVE.
31. A parallel plate capacitor, consisting of two closely spaced metal plates of large area, has a
capacitance of 1 F, and is charged to a voltage 2 V. If the voltage is doubled to 4V and the
distance between the plates is halved, what happens to the energy stored in the capacitor?
(A) stays the same
(D) increases by a factor 8
(B) increases by a factor 2
(E) increases by a factor 16
(C) increases by a factor 4
enegy = ½ C V^2, and we know C depends INVERSELY with plate distance.
So if distance is halved, C goes UP by 2. And if V doubles, V^2 goes up by 4.
So we have 2 from C, and 4 from V^2 = a total increase of 8
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32. A long solenoid with many turns per length has
a uniform magnetic field B within its interior.
Consider the imaginary rectangular path of length c
and width a with one edge entirely within the
solenoid, as shown.
r r
What is ! B " dl around this rectangular path in the
c
B
counterclockwise direction?
A) B c
D) 0
B) 2 B (c + a)
E) B a c
a
C) 2 B c
There is no B field OUTSIDE, and any B on the vertical segments will be
perpendicular to the segments, so the only contribution is along the horizontal path
INSDE the solenoid. There, B is constant, so the integral is simply B*c, that's it.
33. What is the equivalent resistance of the
a
following network of resistors (that is the resistance
between points a and b?
A) 6!
D)
6
!
11
11
B) !
5
E)
5
C) !
6
2!
1!
11
!
6
3!
2 and 3 are in series with each other, R23 = 5 Ohm
We have 1 Ohm in parallel, with 5 Ohm, the result is
1/R(eff) = 1/1 + 1/5 = 6/5. That's the INVERSE of
the answer, which must be 5/6 Ohm.
b
a
34. What is the equivalent capacitance of the
following network of capacitors (that is the
capacitance between points a and b?
A) 6 F
D)
6
F
11
B)
E)
11
F
5
C)
5
F
6
2F
1F
11
F
6
b
Capacitors add "backwards" from how resistors do,
remember? So 2 and 3 add as inverses, 1/R(eff,2-3) = ½ + 1/3 = 5/6.
Thus, R(eff,2-3) = 6/5. Then you ADD this to 1 F to get 1+6/5 = 11/5 F.
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3F
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The next two questions refer to this circuit. A
student chooses the directions of the currents as shown.
35. Which of the following equations is a correct
current equation (only one is correct)?
(A) I1 + I2 = I3
(B) I1 + I3 = I2
(C) I2 + I3 = I1
(D) I1 + I2 + I3 =0
I3
I1
V1
R1
I2
V2
R2
R3
This is Kirchhoff's current law, total current
INTO the node at the top middle = total
current OUT. In this case, I1+I3 are in, and I2
are going out.
36. Which of the following equations is a correct voltage equation (only one is correct)?
(A) V1 +V2– I3R3 = 0
(B) V1 - I2R2 – I1R1 + I1R3=0
(C) V1 + I2R2 + I1R1 - I1R3 = 0
(D) V2 – I2R2 = 0
This is Kirchoff's voltage law. Just watch your signs. If you pick the small loop on the right side,
and go around it counterclockwise, starting at the bottom, you get +V2 - I2R2 = 0. (If you go the
other way, you get -V2 + I2R2=0, which is the same equation!) None of the others is correct,
they all have a sign error on one of the terms! (Actually, A doesn't correspond to any loop at all.
In B, the I1R3 term has the wrong sign, because I1 is going LEFT through R3. In C, the I1R1
and I2R2 terms are wrong.
37. A radio tower is emitting an electromagnetic wave in all directions (isotropically). What is
the ratio of the intensity at 100m to the intensity at 400m?
[Ignore reflections, that is assume the wave is absorbed by any surface it hits]
A) 1
B) 2
C) 4
D) 16
EI measured
at 100m
E) 256
EI measured
at 400m
Intensity = P0/(4 pi r^2), it drops off like 1/r^2. So if you are 4 times closer, you will be 4^2 = 16
times more intense.
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16
38. In case (I) a positive charge +Q is surrounded by a
spherical Gaussian surface of diameter R. In case (II), two
charges, each +Q, are enclosed by a spherical Gaussian
surface of diameter 2R. How does the electric flux through
the surface in case (II) compare with the electric flux
through the surface in case (I)?
A) the same
B) larger by a factor 2
C) larger by a factor 4
D) larger by a factor 8
+Q
(I)
+Q
(II)
+Q
R
2R
E) larger by a factor 16
Gauss' law says total flux = Q(enc)/epsilon0. The radius
(diameter) doesn't matter!
If you enclose twice the charge, you have twice the total flux!
39. A coil of wire carrying current I can rotate freely about
an axis in a uniform magnetic field. The coil is in the plane
of the page and the magnetic field is directed from left to
right as shown. If released from rest in the position shown,
which way does it rotate?
(A) the right side will move out of page
(B) the left side will move out of page.
(C) the loop will not rotate at all
Use the right hand rule. The left side has a force OUT of the
page, the right wire has a force INTO the page.
B
I
40. A coaxial cable consists of an inner solid wire with
radius R and an outer cylindrical wire with inner radius 2R
and outer radius 3R . The inner wire carries total current I
into the page and the outer cylindrical wire carries the same
magnitude current I out of the page. Both wires have uniform
current density. What is the magnitude of the magnetic field B at
the surface of the outer wire, that is, at r = 3R?
A) zero B)
µo I
2! R
C)
µo I
!R
D)
3µ o I
2! R
axis
R
I (in)
2R
3R
I (out)
E) None of these.
Ampere's law says {integral of B around a loop) = mu0*(Total I enclosed). Here, we have I(in)
and I(out), and they add to ZERO total I enclosed! If the integral is zero, by symmetry, B=0 all
the way around. This is the main reason why you buy coax cables for electronic equipment (like
your TV-VCR-DVD connections), so there won't be any stray magnetic fields!
Have a great holiday break, everyone! We had fun teaching you, hope you enjoyed the course.
Vers.0001