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PH202-NG Test 2 (July. 7, 2009, 3:00PM-5:05PM)
• You may not open the textbook nor notebook.
• A letter size information may be used.
• A calculator may be used. However, mathematics or physics formula programmed in a calculator may not be used.
• Write down, reasoning, calculation and answer in the blank space after each problem. Use
the backside of the sheet if necessary.
• Answers without reasonable explanation or convincing mathematical derivation will receive
no point even if your answers coincide with the correct ones. Your reasoning must begin with
basic physical principles. Substantial partial credit may be given to correct reasoning and
mathematical procedures when your final answers are wrong. However, if your final answers
are too obviously wrong, a partial credit may not be given.
Important Physical Constants:
• electron’s electric charge: e = −1.60 × 10−19 C
• electron’s mass: me = 9.11 × 10−31 kg
• proton’s mass: mp = 1.67 × 10−27 kg
• permittivity of free space: ǫ0 =
k = 8.99 × 109 N · m2 /C 2
1
4πk
= 8.85 × 10−12 C 2 /(N · m2 ) where
• permeability of free space: µ0 = 4π × 10−7 T m/A
• acceleration due to gravity at the surface of the earth: g = 9.80 m/s2
• speed of light in a vacuum‘: c = 2.99792458 × 108 m/s
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Total
1. [10 pts.] A proton is projected perpendicularly into a magnetic field with a certain velocity
and follows a circular path. Then an electron is projected perpendicularly into another magnetic
field with the same velocity. The electron follows the exact same circular path as the proton. The
magnetic field for the proton has a magnitude of 0.60 T in the +z direction. Find the direction
and magnitude of the magnetic field for the electron.
To form a circular motion in the same direction, the direction of magnetic force exerted on the
proton must be the same as that on the electron. Since electron and proton have the opposite
sign of electric charge, the direction of the magnetic force on the electron must be opposite to
that on proton. −z direction
To form a uniform circular motion, the magnitude of the magnetic force must match to the
centripetal force.
mv 2
mv
= evB → r =
r
eB
Plug in the masses and magnetic fields for the proton and the electron,
rp =
mp v
eBp
re =
me v
.
eBe
Since the radius of the circle must be the same for the proton and the electron.
re = rp
→
mp v
me v
=
eBe
eBp
→
mp
me
=
Be
Bp
Hence,
Be =
me Bp
(9.11 × 10−31 )(0.60)
=
= 3.3 × 10−4 T
mp
1.67 × 10−27
2
2. [20 pts.] Two long, straight wires carry an identical identical current of I1 =I2 =2.5 A as shown in
Figure. A 4.0-µC point charge is moving at 35 m/s
along the wires in direction B shown in Figure. The
distance between the wires are 6.0 cm and the distance between wire 1 and the charge is 4.0 cm. Answers the following questions. The direction should
be specified by a letter (U , D, F , B, L, R) shown
in the figure.
(a) Find the magnitude and direction of the magnetic field at the position of the charge.
(b) Find the magnitude and direction of the force
exerted on the charge by the magnetic field
(a) Each wire generates circular magnetic fileds around it as shown
in Figure. At the position of the charge, B1 is upward and B2
downward. Theie magnitudes are
B1 =
B2 =
µ0 I
4π × 10−7 2.5
=
= 1.25 × 10−5 T
2πr1
2π0.040
4π × 10−7 2.5
µ0 I
=
= 2.50 × 10−5 T
2πr2
2π0.020
Since B2 is larger than B1 , the net magnetic field is downward
D.
The magnitude of net magnetic field is
Bnet = |B1 − B2 | = |1.25 × 10−5 − 2.50 × 10−5 | = 1.3 × 10−5 T
(b) Since the velocity and the magnetic field are in direction B and D, respectively, RHR-1
indicates that the direction of force on the charge is L
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3. [20 pts] Figure shows a rectangular N = 20-turn coil of
wire (ABCD). Its dimensions is BC = 12 cm by AB = 6.0 cm.
It is mounted in the xy plane and hinged along one long
side (AD) on the y axis. It carries a current of i = 0.10 A.
The direction of the current is shown in Figure. A uniform
magnetic field of magnitude B = 0.50 T , is applied at 30◦
from the x axis as shown in Figure.
(a) Find the direction and magnitude of the force exerted
on the wire AB. The direction should be expressed like
+x.
(b) Which edges (AB, BC, CD, DA) contribute the torque
around the hinge. List all of them.
(c) Find the magnitude of the torque. You must derive it
from basic physics principles.
(d) In which direction does the coil rotate, viewed from the
top, clockwise or counterclockwise?
(a) Using RHR-1, the direction is −y . The magnitude is given by
F = N IBL sin θ = (20)(0.10A)(0.50T )(0.060m) sin 30◦ = 0.030 N .
(b) The force on the segments AB and CD are in ±y dircetion which is parallel to the axis of
rotation. Hence, the force does not contribute the torque. The force on DA has zero lever arm.
Hence, it has no contribution to the torque. Only BC generates noz-ero torque.
(c) Figure shows the direction of magnetic field, force and lever
arm. The magnitude of the magnetic force on BC is
F = N IBL sin φ = (20)(0.10A)(0.50T )(0.12m) sin 90◦ = 0.12 N
and the lever arm
ℓ = (0.060m) cos 30◦ = 0.052 m
Hence, the torque is
τ = ℓF = (0.052m)(0.12N ) = 6.2 × 10−3 N · m
(d) As shown in Figure, the loop rotates in clockwise direction.
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4. [20 pts.] A metal rod of length 15 cm on two parallel
metal tracks is pulled with a hand to the right at a constant
speed of 1.2 m/s. The tracks are connected at one end so that
they and the rod form a closed circuit as shown in Figure.
The rod has a resistance 10 Ω, and the tracks have negligible
resistance. A uniform magnetic field of 2.0 T perpendicular
to the plane of this circuit (into paper) is applied.
(a) What is the magnitude and direction (up or down) of the induced current in the rod? (The
direction should be one of “Up”, “Down”, “Left”, “Right” or “No current”.)
(b) Find the magnitude of the force exerted on the rod by the hand.
(a) Using RHR-1, the direction is UP .
Emf generated in the loop is given by emf = vBL. Using the Ohm’s law
I=
vBL
(1.2m/s)(2.0T )(0.15m)
emf
=
=
= 3.6 × 10−2 A .
R
R
10Ω
(b) Since the velocity is constant, Fnet = Fhand − Fmag = 0. Hence,
Fhand = Fmag = IBL = (3.6 × 10−2 A)(2.0T )(0.15m) = 1.1 × 10−2 N .
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5. [20 pts.] A long and narrow rectangular loop of wire
is moving toward the bottom of the page with a speed of
0.025 m/s (see the drawing). The loop is leaving a region
in which a 3.4-T magnetic field (out of paper) exists. The
magnetic field outside this region is zero.
(a) During a time of 2.0 s, what is the magnitude of the
change in the magnetic flux?
(b) Find the magnitude of emf.
(c) What is the direction of induced magnetic field, out of
paper or into paper? Justify your answer.
(d) What is the direction of induced current in the loop,
Clockwise or coounterclockwise?
(a) Durin the time of falling, the loop moves down by d = vt. Hence the area of the region with
magnetic field decreases by ∆A = −vtℓ. (The negative sign is due to the decreasing are. Hence,
the change in the magnetic flux is
|∆Φ| = |B∆A| = Bvtℓ = (3.4T )(0.025m/s)(2.0s)(0.080m) = 1.4 × 10−2 wb .
(b) Using the Faraday’s law
|emf| =
|∆Φ|
= Bvℓ = (3.4T )(0.025m/s)(0.080m) = 6.8 × 10−3 V .
t
(c) Since the oroginal magnetic field inside the loop is decreasaing, the Lenz’s law indicates that
the induced magnetic field should be in the same direction as the direction of the original
magnetic field. Out of paper .
(d) In order to induce the out-of-paper magnetic field, the induced current flows in the
counterclockwise direction.
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6 [10 pts.]
(a) Which of the following electromagnetic waves have the shortest wave length in a vacuum?
Visible light
X-ray
Gamma-ray
Microwave
Ultra violet
(b) Which of the following electromagnetic waves have the highest frequency in a vacuum?
Visible light
X-ray
Radiowave
Infra red
Gamma-ray
(c) Obtain the wavelength in a vacuum for red light, whose frequency is 4.22 × 1014 Hz.
λ=
3.00 × 108 m/s
c
=
= 7.1 × 10−7 m .
f
4.22 × 1014 Hz
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7. Extra [15 pts.] (Difficult)
A conducting rod slides down between two frictionless vertical
copper tracks at a constant speed of 4.0 m/s perpendicular to
a 0.50-T magnetic field. The resistance of the rod and tracks
is negligible. The rod maintains electrical contact with the
tracks at all times and has a length of 1.3 m. A 0.75-Ω resistor
is attached between the tops of the tracks.
(a) What is the mass of the rod?
(b) Find the energy dissipated in the resistor while the rod
falls for 0.20 s.
(c) The induced current dissipates energy in the resistor.
Since energy must conserve, the same amount of energy must come from another form of energy. What is
it? Show that the amount of energy lost during 0.20 s
matches to the dissipation obtained in part (b).
(a) Since the velocity is constant, Fnet = Fmag − mg = 0. Therefore, m = Fmag /g. The magnetic
force is due to the current induced by the change in magnetic flux insize the loop. The current is
given by
vBL
emf
=
I=
R
R
Hence, the magnetic force given by
¶
µ
vB 2 L2
vBL
BL =
Fmag = IBL =
R
R
Thus,
m=
(4.0m/s)(0.50T )2 (1.3m)2
vB 2 L2
=
= 0.23 kg .
Rg
(0.75Ω)(9.8m/s2 )
(b)
E = P t = Iemft =
(emf)2 t
(vBL)2 t
[(4.0m/s)(0.50T )(1.3m)]2 (0.20s)
=
=
= 1.8 J .
R
R
0.75Ω
(c) During the process, the height of the rod is lowered. Then, the
potential energy of the rod due to gravity is reduced. The amount of the lost potential energy
appears as the energy dissipated in the resistor. In fact, the loss of the potential energy
mgh = mgvt = (0.23kg)(9.8m/s2 )(4.0m/s)(0.20s) = 1.8 J
which agrees with the value obtained in part (c).
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8. Test 1 Bailout Problem [Maximum 20 pts.] Do not solve this problem if your test 1 score
was higher than 70. The points from this problem will be added to your score of test 1. However,
your test 1 score will not exceed 70 pts. For example, if your score was 60 pts, the maximum point
you can earn from this problem is 10 pts.
A 120-V battery and four 60-Ω light bulbs are connected as shown in
Figure.
(a) [5 pts.] How much current is drawn from the battery?
(b) [5 pts.] How much power is consumed by the whole system?
(c) [5 pts.] Find the power dissipated in the light bulb L4 .
(d) [5 pts.] If the light bulb L3 burns out, does the light bulb L1
become brighter or darker. Justify your answer.
(a) Since L2 , L3 , and L4 are in parallel connection their equivalent registance is
1
1
1
1
1
1
1
1
=
+
+
=
+
+
=
R234
R2 R3 R4
60Ω 60Ω 60Ω
20
→
R234 = 20 Ω
Now, R1 and R234 are in series, the net equivalent resistance is Req = R1 + R234 = 80 |omega.
Using the Ohm’s law,
V
120V
I=
=
= 1.5 A .
Req
80Ω
(b)
P = IV = (1.5A)(120V ) = 180 W .
(c) The voltage across L1 is V1 = IR1 = (1.5A)(60Ω) = 90V . Then, the voltage accross L4 is
given by V4 = V − V1 = 120V − 90V = 30V . Hence the power consumed by L4 is
P4 = I4 V4 =
v42
(30V )2
=
= 15 W .
R4
60Ω
(d) When L3 burns out, the equivalent resistor of L2 and L4 becomes larger than R234 obtained
in part (a). Hence, the voltage accross L1 decreases. Then, the power consumption of L1
decreases. L1 gets darker .
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