Download PH202-5D Test 2 (July 11, 2007)

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PH202-5D Test 2 (July 11, 2007)
• 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 nor mathematical derivation will receive no point. Partial
credit may be given to correct reasoning and mathematical procedures even if your final answers are
wrong.
• Don’t forget units in the final answers!
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
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permittivity of free space: ²0 = 4πk
= 8.85 × 10−12 C 2 /(N · m2 ) where k = 8.99 × 109 N · m2 /C 2
permeability of free space: µ0 = 4π × 10−7 T · m/A
[a]
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1.1 [3 pts.] Find physical quantities that depend on the color of light from the followings. There may be
more than one (No reasoning is required.)
(a) Speed of light
(b) Wave length
(c) Brightness
(d) Frequency
(e) Polarization
Wave length and Frequency
1.2 [3 pts.] Order the following electromagnetic waves in terms of their wave length from short to long
wave length.(No reasoning is required.)
(a) X-ray
(b) Ultraviolet
(c) Gamma ray
(d) Visible light
(e) Microwave
(f) Infrared
(c) – (a) – (b) – (d) – (f) – (e)
1.3 [3 pts.] There are two types of doped semiconductors; one with mobile electrons is called p type and
the other with mobile holes n type. A voltage is applied to a p − n junction as shown in Figure. Which one
allows much larger electric current than the other? Briefly, justify your answer.
Figure B
Figure A
Figuare 1A shows forward bias and 1B shows reverse bias. Therfore, Figure 1A allows larger current.
Reason: In forward bias, holes in p-type and electrons in n-type are driven by the battery toward the
intereface of the p-n junction where they are combined. At the same time, the battery provides p-type with
holes and n-type with electrons. As a results, continuous current flows. On the other hand, in reverse bias,
holes are electrons are driven away from each other by the battery generates an electric potential across the
junction which opposes to the battery. Therefore, practically no current flows.
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2. [15 pts.] The triangular loop of wire shown in the drawing
carries a current of I = 5.0A. A uniform magnetic field is directed
parallel to side AB of the triangle and has a magnitude of 2.0 T as
shown in Figure 2. In the following questions, the direction should
be expressed as “left”, “right”, “up”, “down”, “into page”, or “out
of page”.
(a) Find the magnitude and direction of the magnetic force exerted on the side AB.
(b) Find the magnitude and direction of the magnetic force exerted on the side BC.
(c) Find the magnitude and direction of the magnetic force exerted on the side CA.
Figure 2
(a) Since the magnetic field and current are parallel to each other, no force is exerted on the edge AB.
Magnitude = 0 N; No direction (b) The angle between the magnetic field and current is θ=55◦ . The length
of edge BC is LBC = LAB / cos 55◦ .
FBC = IBLBC sin θ = 5.0 × 2.0 ×
2.0
× sin 55◦ = 20 tan 55◦ = 29
cos 55◦
Magnitude: 29 N; Direction: out of page
(c) The angle between the magnetic field and current is θ=90◦ . The length of edge CA is
LCA = LAB tan 55◦ .
FCA = IBLCA sin θ = 5.0 × 2.0 × 2.0 tan 55◦ × sin 90◦ = 20 tan 55◦ = 29
Magnitude: 29 N; Direction: into page
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3. [20 pts.] Figures 3A and 3B show a bar magnet falling through
a metal ring. (Don’t forget to justify your answers.)
(a) Find the direction of the induced current when the magnet
is still above the ring? Draw an arrow with a symbol ”I” in
Figure 3A. Explain your answer. (Don’t forget justification).
Figure 3A
The magnet is falling above the ring.
(b) When the magnet is below the ring, draw induced magnetic
field lines in Figure 3B. Explain your answer.
(c) Explain why the motion of the magnet is retarded when the
magnet is above the ring.
(d) In which direction is the magnetic force exerted on the magnet by the current in the ring when the magnet is below the
ring? Explain your answer.
(a) As the magnet falls, downward magnetic field increases inside
the ring. Then, upward magnetic field is induced (Lenz’s law). In
order to induce such magnetic field, a current flows as shown in
Figure 3A.
(b) As the magnete falls, downward magnetic field decreases inside the ring. Acording to Lenz’s law, downward magnetic field is
induced inside the ring as shown in Figure 3B.
(c) The induced magnetic field opposes to the magnetic field of the
magnet. Therefore, the ring exerts a repulsive force on the magnet,
which reduces the downward acceleration due to gravity.
(d) The induced magnetic field is in the same direction as the magnetic field of the magnet inside the ring. Then, the ring exerts an
atractive force on the magnet. Therefore, the direction of the force
is upward.
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The magnet is falling below the ring.
Figure 3B
4. [20 pts.] A long, straight wire carries a current of I = 2.0A. A
−5.0 µC charge is moving at 30 m/s along the wire in direction
F shown in Figure 4. The distance between the wire and charge is
2.0 cm.
Figure 4
(a) Find the magnitude and direction of the magnetic field at the
position of the charge. The direction should be specified by
a letter (U , D, F , B, L, R) shown in the figure.
(b) Determine the magnitude and direction of the force exerted
on the charge by the magnetic field The direction should be
specified by a letter (U , D, F , B, L, R) shown in the figure.
(a)
B=
µ0 I
4π × 10−7 × 2.0
=
= 2.0 × 1005
2πr
2π × 0.02
Direction of the magnetic field is determined by righthand rule. (The red arrow in the Figure 4b.)
Magnitude: 2.0 × 10−5 T; Direction: D
(b)
F = |qvB| = 5.9 × 10−6 × 30 × 2.0 × 10−5 = 3.0 × 10−9
Using right-hand rule and taking into account the negative
sign of the charge, the charge is attracted to the wire.
Magnitude: 3.0 × 10−9 N; Direction: L
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Figure 4a
5. [18 pts.] A proton (e = 1.60 × 10−19 C and mp = 1.67 × 10−27 kg) moves perpendicular to a uniform
magnetic field. Its trajectory is a circle of radius r = 0.25m with a period of T = 2.1 × 10−7 s.
(a) What is the speed of the proton?
(b) What is the centripetal force required for the proton to make such a circular motion?
(c) Find the magnitude of the magnetic field. (You must derive the answer from basic principles.)
(a)
v=
2πr
2π × 0.25
= 7.48 × 106
=
T
2.1 × 10−7
speed = 7.5 × 106 m/s
(b)
Fc =
mv 2
1.67 × 10−27 (7.48 × 106 )2
=
= 3.74 × 10−13
r
0.25
centripetal force = 3.7 × 10−13 N
(c) The centripetal force is provided by the magnetic force. Therefore,
evB =
mv 2
r
Solving this equation for B,
B=
mv
1.67 × 10−27 × 7.48 × 106
=
= 0.312
er
1.60 × 10−19 × 0.25
the magnitude of the magnetic field = 0.31 T
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6. [18 pts.]Figure 6 shows a copper wire (negligible
resistance) bent into a circular shape with a radius of
r = 0.45m. The radial section BC is has resistance
R = 2.5Ω and fixed in place, while the copper bar AC
(negligible resistance) sweeps around at an angular speed
of ω = 12rad/s. The bar makes electrical contact with
the wire at all times. A uniform magnetic field exists everywhere, is perpendicular to the plane of the circle, and
has a magnitude of B = 3.2 × 10−3 T .
Figure 6
(a) Find change in the magnetic flux during t = 0.1s.
(b) Find the magnitude of the current induced in the
loop.
(c) Is the direction of the induced current clockwise or
counterclockwise?
(a) Te area inside the closed loop is a = 12 r2 θ where θ is the angle BCA. Since the bar AC moves with a
constant angular velocity, θ = θ0 + ωt. Tnen the change of the are is
1
1
∆a = af − ai = r2 (θ − θ0 ) = r2 ω∆t
2
2
Hence, the change in the magnetic flux is
1
1
∆Φ = B∆a = Br2 ω∆t = 3.2 × 10−3 × (0.45)2 × 12 × 0.1 = 3.9 × 10−4
2
2
∆Φ = 3.9 × 10−4 Wb
(b)
emf = −
∆Φ
1
1
= − Br2 ω = − 3.2 × 10−3 × (0.45)2 × 12 = −3.9 × 10−3
∆t
2
2
|I| = |
3.9 × 10−3
emf
|=
= 1.6 × 10−3
R
2.5
|I| = 1.6 × 10−3 A
(c) Since the magnetic field goes into page and the magnetic flux increases, the induced magnetic field is
out of page. Therfore, the current is counterclockwise.
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