Reitze/Kumar
PHY2054
Exam 2
March 23, 2011
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Useful Constants:
ke= 8.99×109 Nm2/C2
ε0= 8.85×10-12 C2/(Nm2)
µ0 = 4π × 10-7 Tm/A
k0 = µ0/(4π) = 10-7 Tm/A
electron charge = -e = -1.6×10-19 C electron mass= 9.11×10-31 kg
V=Volt N=Newton J=Joule m=Meter C=Coulomb g = 9.8 m/s2 c = 3×108 m/s
"milli-" = 10-3 "micro-" = 10-6 "nano-"=10-9 "pico="=10-12
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1. A magnetic field CANNOT:
A magnetic field does not change the speed of a moving charged particle only its
direction. It applies a force on a moving charged particle. It changes the velocity
(direction), the momentum as well as the trajectory of motion.
(1) change the kinetic energy of a charged particle (2) exert a force on a charged particle
(3) change the velocity of a charged particle (4) change the momentum of a charged
particle (5) change the trajectory of a charged particle
2. A proton (charge +e), traveling perpendicular to a magnetic field, experiences the
same force as an alpha particle (charge +2e) which is also traveling perpendicular to the
same field. The ratio of their speeds, vproton/valpha, is:
F = q1v1B = q2v2B or v2/v1 = q1/q2 = 2
(1) 2
3. An infinitely long straight wire and a single circular curent loop
with radius R are arranged as shown in the figure. If both the straight
wire and the loop are carrying the same current I, what is the
magnitude of the magnetic field at the center of the loop?
I
I
R
There is a log wire and there is a loop. The field at point P (loop center) due to the wire
is in the same direction as that due to the loop, coming out of the plane of the paper. The
field due to the wire is µI/2πR and that due to the loop is µI/2R. The answer follows
(1) (π+1)µ0I/(2πR)
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I
4. An electron (charge q = -1.6 ×10-19 C) undergoing uniform circular motion with a
radius R = 2 ×10-8 m acts like a current loop. If the speed of the electron is 20 km/s what
is the corresponding current flow around the circle (in Amps)?
I = e v/2πR = 25 nA.
(1) 2.5×10-8
5. Two isotopes of helium start from rest and pass through a potential
difference ∆V = (860, 480) V and then move out of a slit at the point S into
a (0.2, 0.4 T uniform magnetic field pointing into the paper as shown in the
figure. Helium-4 consists of two protons and 2 neutrons (atomic number 2,
atomic mass = 4 AU) and helium-3 has two protons and one neutron
(atomic number 2, atomic mass = 3 AU). If both isotopes are doubly
ionized (i.e. both have charged +2e) what is the distance of separation (in
mm) of the two isotopes when the reach the photographic plate P? (Note: 1 AU =
1.67×10-27 kg)
R = mv/qB = p/qB. The momentum p increases from zero (at rest) to √(2mq∆V). Thus
the distance between the two ions δ = 2(R2-R1) =
= 8mm
For (∆V, B) = (860 V, 0.2T) δ = 8 mm
(860 V, 0.4T)
4 mm
(480 V, 0.2T)
6 mm
(1) 8
6. A loop of wire carrying a current of (2.0, 4.0) A is in the shape of a right
triangle with two equal sides, each with length L = (15, 30) cm as shown in the
figure. A 0.7 T uniform magnetic field is in the plane of the triangle and is
perpendicular to the hypotenuse. The magnetic force on either of the two equal
sides has a magnitude of:
B
L
L
The force on the bottom branch is F = ILB sin45 = ILx.7/√2
For (I,L) = (2A, 15 cm)
F= 0.15 N
(2A, 30 cm)
0.30 N
(4A, 15 cm)
0.30 N
Can you solve the problem where the sides are not equal and the angles (instead of 45)
are (60, 30)? Assume that the hypotenuse is 15 cm long.
(1) (0.15, 0.3) N
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7. In the circuit shown in the figure, L = 56 mH, R = 4.6 Ω and V = 12.0 V. The
switch S has been open for a long time then is suddenly closed at t = 0. At what
value of t (in msec) will the current in the inductor reach (1.1, 2.1, 0.57) A?
which can be solved for t for a given I,
(1) (6.67, 19.9, 3.0)
8. In the previous problem, what is the total energy stored in the inductor a long time
after the switch is closed?
Once everything settles down, the current I = V/R and in the inductor, the energy E=
½LI2.
(1) 0.19 J
9. As shown in the figure, four wires have current flowing perpendicular to
the plane of the page. The wires go through the corners of a square. If wires
1 and 4 each have current I flowing out of the page, and wires 2 and 3 each
have current I flowing into the page, what is the direction of the magnetic
field at the center of the square?
The magnetic field at the center due to the coming out current in wire 1 is
pointed towards wire 2. That due to the coming out current in wire 4 is
pointed towards wire 1. That is also the direction of the magnetic field due to the
ingoing current in wire 2. The ingoing current in wire 3 gives rise to a magnetic field in
the direction pointed towards wire 2. All currents are equal, all distances are the same.
The net magnetic field is a vector sum in the direction positive y.
(1) positive y
10. A uniform magnetic field is directed into the page. A charged particle
moves in the plane of the page and follows a clockwise spiral of decreasing
radius as shown in the figure. Which of the following statements is true?
Watch the arrow. The velocity is pointed to the right. The force vxB would
be directed up but the particle moves down. It must be negatively charged. The circle is
spiraling in to smaller values and r = mv/qB so v must be slowing down.
(1) the charge is negative and slowing down
11. The distance from earth to a NASA satellite traveling through the solar system is (6.0,
3.0, 0.6) × 109 m. A command is sent to the satellite from scientists on earth to send back
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telemetry data. Assuming the satellite instantly transmits the data when it receives the
command, how much time (in s) would pass before the scientists receive the data?
The time spent in the round trip = 2L/c.
(1) (40, 20, 4)
12. A 10(N) turn circular loop with radius 0.01 m (A = πr2) and
resistance 5 Ω (R) is placed in the center of a solenoid. The
solenoid has a length 0.8 m (L), a radius of 0.1 m, and 1000 turns
(nL). The current in the solenoid is ramped up uniformly from 2
A to (12, 22) A in (1, 2) s. What is the average current (in µA) in the loop during this
time?
For ∆I/∆t = (10, 20) A/s
I = (π2, 2π2) µA
(1) (9.9, 19.7)
13. The normal to a certain 1.0 m2 area makes an angle of 60o with a uniform magnetic
field. The magnetic flux through this area is the same as the flux through a second area
that is perpendicular to the field if the second area is:
When the area is perpendicular to the magnetic field, the normal to the area is parallel.
Φ = BA1cos 60 = BA2
A2 = 0.5
(1) 0.5 m2
14. A conducting bar with mass M = 10 kg can slide with no friction along
two conducting rails separated by distance L = 0.5 m, as shown in the
figure. A uniform 2 Tesla magnetic field points into the page, as indicated
by crosses. An external force Fapp pulls the bar to the right with a constant
speed v = 3 m/s. What is the magnitude and direction of the induced
current in the R = 0.5 Ohm resistor?
. The current direction should, following Lenz’ law, produce
a field out of the plane of the paper. i.e counter-clockwise.
(1) 6 Amps counter-clockwise
15. In the previous problem, what is the magnitude of the external force Fapp (in N)?
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Since the motion of the rod is at a constant velocity, unaccelerated, the applied force
must be the same as the magnetic force. That is F = BLI = (BL)2v/R.
Another way to get that result is to calculate the power I2R in the circuit and equate that
to vFapp as the power supplied. The answer is the same.
(1) 6
θ
16. A light ray enters a rectangular glass slab (n = 1.5) at point A at an
incident angle θ and then undergoes total internal reflection at point B at the
interface between the glass and water (n = 1.3) as shown in figure. What is the
maximum incident angle θ (in degrees) for this to occur?
Air
At the water-glass interface, the angle of incidence, call it θ1, is related to the angle of
refraction at the air-glass interface, the latter being 90- θ1. Total internal reflection at
the water glass interface means that 1.3 = 1.5 sin θ1. At air-glass interface, sin θ = 1.5
cos θ1 or
(1) 48.4
(2) 31.3
(3) 23.6
(4) 41.5
(5) 62.8
17. Maxwell’s equations predict that the speed of light in free space is
The important bit here is to realize that all other answers (except the first one,
independent of frequency) are wrong. Answers 2,3 are not true in free space. In a
material speed of light depends on frequency and that fact is derivable from Maxwell’s
equations.
(1) independent of frequency
(2) an increasing function of frequency
(3) a decreasing function of frequency
(4) a function of the distance from the source
(5) a function of the size of the source
18. The eye is most sensitive to light of wavelength 550 nm, which is in the green-yellow
region of the visible electromagnetic spectrum. What is the frequency (in Hz) of this
light?
f = c/λ
(1) 5.45×1014
(2) 5.45×1013
(3) 5.45×1015
(4) 1.82×1014
A
Water
n = 1.3
(5) 1.82×1013
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B
Glass
n = 1.5
19. Assume the solar radiation incident on Earth is 1.34 kW/m2 (at the top of the Earth’s
atmosphere). Calculate the total power radiated by the Sun (in W), taking the average
separation between the Earth and the Sun to be 1.49×1011 m.
Total power irradiated by sun = 1340 x 4π (1.49x1011)2
(1) 3.74×1026
(2) 4.48×1026
(3) 2.24×1026
(4) 3.74×1024
(5) 4.48×1024
20. A speeder tries to explain to the police that the yellow warning lights on the side of
the road looked green to her because of the Doppler shift. How fast (in km/s) would she
have been traveling if yellow light with a wavelength of 580 nm (λs) had been shifted to
green with a wavelength of 560 nm?
fo = c/λo = c/ λs [1+u/c] thus u = c[λs/λo -1] = 10, 714 km/s
(1) 10,700
(2) 21,400
(3) 5,300
(4) 1,700
(5) 700
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