Tests Concept -- Final Review CTF-1. A charge q is released from rest at point in empty space were there may be E- and/or B-fields. There are no forces on the charge except for the forces due to the E and/or B-fields (no gravity, etc.). The charge is observed for a short while and is seen to move along a curved path. Which one statement must be true? A: There might be only a (non-zero) E-field present and no B-field. B: There might be only a (non-zero) B-field present and no E-field. C: There must be both a non-zero E-field and a non-zero B-field present. D: None of the above statements must be true/don't know.
q
Answer: This is a very tricky problem (trickier than what I would give on the Final). You have to distinguish carefully between the words "might" and "must". The correct answer is: There might be only a (non-zero) E-field present and no B-field. If the particle is in a curved E-field, it will accelerate along a curved path. It is true that this path could be caused by a B- and E-field acting together, but it is not true that the path must be caused by a B- and E-field together.
CTF-2 Points A and B are distances r and 2r respectively from a point charge q. What is the voltage difference V = VB - VA ?
r q
A
r
B
A:
kq r
B:
1 kq 2 r
C:
3 kq 2 r
D: None of these.
Answer: None of these. Again, a sneeky, tricky answer, which I would not do on the kq kq 1 kq - =- . The minus sign matters. If q is positive, Final. V = VB - VA = 2r r 2 r then V is negative; if q is negative, then V is positive. The same formula works regardless of whether q is positive or negative.
CTF-3 A charge q is located at the center of an gaussian cubical box as shown. A student is asked: What is the electric flux E though the top face of the cube? Is the answer hard to calculate or easy to calculate? A: Easy B: Hard.
q
Answer: Easy. The total flux through the box is q/o. By symmetry, the flux through one face is 1/6 th the flux through the whole box. Flux = (1/6) q/o.
CTF-4 A capacitor with capacitance C is attached to a battery with voltage V. What is the flux r r E da through the cubical volume shown? (The end faces of the cube are within the
z
metal plates of the capacitor.
C
A:
CV o
B:
2CV o
C: zero
D: None of these.
Answer: Zero. In electrostatic situations, the E-field inside a conductor is zero. Since E=0 on the right and left ends of the cubical r r (imaginary) box, the flux through those ends is zero. On the sides of the box, E da = 0 , r r since Eda . So the total flux is zero.
V
Since
z
r r E da = 0 , then by Gauss's Law, the total charge must be zero within the box
must be zero, so we have proven that the two plates must have equal and opposite charge densities. CTF-5 A resistor is plugged into a 120VAC wall socket. The graph below is either voltage V across, current I in, or power P dissipated in the resistor vs. time.
?
time
What could the graph be? A: V only B: I only C: V or I only D: V, I, or P E: something else Answer: V or I only. The power P = IV = I2R is always positive.
?
P
time
I or V
CTF-6 A capacitor has a voltage V across its plates. An electron, initially at rest, is released from a point very close to the negative plate of a capacitor and it accelerates toward the positive plate. The electron has charge -e and mass m. There is no gravity in this problem, and you may assume that the motion of the electron is non-relativistic. What is the final speed of the electron just before it collides with the positive plate? A:
2eV m
2eV m
B:
m 2eV
C:
D: m e V
E: None of these.
V
Answer:
2eV m
KE = PE =
1 mv 2 = e V 2
CTF-7 A square with edge length L is made larger and larger...
(length of perimeter )
As the square grows, the ratio
area
..
A: increases Answer:
B: decreases
C: remains constant.
length 4 L 4 = 2 = . The ratio decrease as L increases. area L L
CTF-8 Calculator question:
105 = ?
A: 1E5
B: 10E5
C: something else.
Answer: 1E5 means "1 times 105" = 105.
CTF-9 When the switch is closed, the current I1 through resistor R1
A: increases, C: stays the same.
B: decreases,
I bat
V
R1
I1
R2
Answer: stays the same. The resistor R1 is attached to the battery with constant voltage V. Regardless of whether the switch is open or closed, the current thru R 1 is I1 = V/R1. What changes when the switch is closed is the total current thru the battery Itot = I1 + I2.
CTF-10 At time to, a particle with charge q has instantaneous velocity v o as shown and
$ in moving in a uniform magnetic field B = B x .
V B q
A: True
or
B: False
As time increases, the x-component of the velocity vx remains constant.
r r r Answer: True, the force F = qv B is into or out of the page (depending on the sign of q), so Fx = max = 0, ax= 0 means vx=constant.
A: True
or
B: False
The of KE the particle remains constant as time increases.
r r True: The magnetic field can do no work on the particle since Fv always. The KE can change only if work is done (by the work-energy theorem: KE = Wnet.)
CTF-11 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.
c a
What is
z
r r B dl ?
A: Bc B: 2B(c + a)
B
C: 2Bc D: None of these
Answer: Bc . Only the bottom of the loop contributes to the line integral. On the sides r r r r B dl , so B dl = 0. On the top of the loop (outside the solenoid), B = 0.
The solenoid has n turns per length and carries a current I. What is
A: o nI
B: o ncI
C: o n( a + c) I
z
r r B dl ?
D: None of these.
Answer: o ncI . By Ampere's Law,
z
r r B dl = o I enclosed = o N I = o nc I .
Here N is the number of turns that thread the loop, n = N/c, N = nc.
CTF-12
A single circular loop of wire of radius R surrounds a long solenoid of radius r, turns per length n, carrying a current I. The diagram shows the end-on view of the situation. What is the magnetic flux through the single wire loop? A: onI R
2
R r solenoid
B: oI r /(2R)
2
I
C: R2(oI/2R) E: None of these.
D: onI r2
Answer: onI r2, NOT onI R2. The B-field is non-zero only within the solenoid, so only the area of the of the solenoid r r contributes to the integral B da .
z
CTF-13 The following questions refer to the circuit shown which has two resistors, both with resistance R, an inductor with inductance L, a battery with constant voltage V, and a switch which can be in position A or B. Suppose the switch has been in position A for a long time and is then switched to position B.
A B V R R L
1. Immediately after the switch is thrown to B, what is the current in the inductor?
A) zero B) V/R C) V/(2R) D) 2V/R E) None of these. Answer: The current thru L just after the switch goes to B must be the same as when the switch was at A: I = V/R
2. Immediately after the switch is thrown to B, what is the voltage across the inductor
L? A: zero B: V C: 2V D: V/2 E: None of these.
Answer: 2V. By Kirchhoff's voltage law, the voltage across the inductor L must be the same magnitude as the voltage across the two resistors: V = I (2R) = (V/R)(2R) = 2V.
CTF-14 A hand-cranked electrical generator is attached to a resistor R. The generator is turned at a constant rate which causes an AC current to flow in the resistor. Suddenly,the value of the resistance R increases. Does the generator get easier or harder to turn?
A: Easier.
B: Harder.
C: No change in difficulty.
Answer: Easier. The power is P = IV (V = emf = constant amplitude AC, since crank is turned at constant rate). Current I = V/R. So P = I V = V2/R. As R increases, the power P decreases. Less power means crank is easier to turn.
emf =V
I
R
CTF-15 An radio tower is emitting a single-wavelength (monochromatic) EM wave in all directions (isotropically). A technician measures the electric field amplitude E as 100 meters from the tower and again 400m from the tower. What is the ratio E100/E400? (Hint: Intensity is proportional to the square of the amplitude: I E2)
E measured at 100m
E measured at 400m
A: 2
B: 4
C: 16
D: 162 = 256
E: None of these.
Answer: 4. Intensity I =
Po 1 1 2 E2 , E 2 4 R R R
CTF-16 A ray of light is sent into an optic fiber at an angle such that the angle with the normal to the pipe is . The fiber has an inner core with index ni and an outer cladding with index no. If no/ni is not large enough there will be no internal reflection. What is the minimum value of no/ni required for total internal reflection to occur?
no ni
A: sin
B: sin-1()
C: 1/(sin)
D: None of these.
Answer: ni sinc = no sin90o = no, (no/ni) = 1/sinc
CTF-17 Is the image on a movie screen real or virtual?
A: real
B: virtual
Answer: Real. When the rays actually converge at a point in space (where you can put a screen to see the image) then the image is real.
Is the image seen with a virtual reality headset real or virtual? A: real B: virtual
Answer: Virtual. The image must be further than 25cm from the viewer's eye for the viewer to be able to focus on the image.
image
CTF-18 A double slit interference pattern is seen on a screen The path difference (p.d.) for a given point on the screen is the quantity [(distance to slit 1) - (distance to slit 2)]. What is the path difference at point A? A
slit 2
d
slit 1
fringes screen
A: 5 B: 3 C: 2 D: None of these.
Answer: None of these. The path difference is 0.5. For the first max off to the side of the central max, p.d.= ; for the second max, p.d. = 2; for the first min off to the side, p.d. = (1/2) ; for the third min off to the side, p.d. = (2.5) .