PHY2054 Spring 2009
Prof. Pradeep Kumar
Prof. Paul Avery
Prof. Yoonseok Lee
Mar. 27, 2009
Exam 3 Solutions
1. Two coils (A and B) made out of the same wire are in a uniform magnetic field with the coil
axes aligned in the field direction. Coil A has 10 | 10 | 20 turns of loop with a radius of 5 | 10 | 5
cm and coil B has 5 turns with a radius of 10 | 5 | 10 cm. Now the field strength is doubled in 2
s. What is the ratio of the emf induced in each coil (emf_1:emf_2)?
Answer: 1:2 | 8:1 | 1:1
Solution: The induced emf in coil i is Ni Ai !" B / !t . The ratio of emfs is thus independent of
2
the field change. Since the area is ! r 2 , the ratio is (N1 A1 ) / (N 2 A2 ) = (N1 / N 2 )(r1 / r2 )
2. An ideal transformer has a 100-turn primary coil and 10,000-turn secondary coil. 1 mA DC
current is flowing through the primary coil and the current suddenly drops to zero in 1 µs. What
is the induced voltage appearing on the secondary coil if the self inductance of the primary coil is
1.2 H | 0.8 H | 1.7 H?
Answer: 120,000 V | 80,000 V | 170,000 V
Solution: The emf in the primary coil is L!i / !t . The emf in the secondary coil is calculated
from the ratio of turns so is equal to 100 L!i / !t .
3. A 100 Ω | 160 Ω | 90 Ω resistor, a 100 µF capacitor, and a 0.45 H inductor are connected in
series to a 120 Vrms 60 Hz power source. What is the average power consumption in this circuit?
Answer: 47.2 W | 50.0 W | 45.4 W
2
R , where the rms current is calculated
Solution: The average power consumption is Pave = I rms
from I rms = Vrms / Z = Vrms / R 2 + X L2 , with X L = ! L = 120" L .
4. A 1.2 | 1.5 | 0.8 kg object is floating near the international space station. The crew members in
the space station direct a 1.2 kW laser perpendicular to a flat surface of the object for an hour.
What is the change of speed of the object? Assume that the object surface absorbs laser light
completely.
Answer: 0.012 m/s | 0.0096 m/s | 0.018 m/s
Solution: The force from the laser is IA / c = P / c , where P is the power of the laser. The change
in velocity Δv is obtained from acceleration × time or !v = at = (F / m )t = Pt / mc .
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PHY2054 Spring 2009
5. As a beam goes through layers of different materials (air, diamond, and ice), it bends as shown
in the figure. The figure indicates the angles formed by each beam and vertical line. When the
angle A (angle of incidence) is 45° | 50° | 40°, what is the angle B (angle of refraction in ice) in
degrees? Note that nair = 1.00, ndiamond = 2.42 and nice = 1.31.
Answer: 32.7 | 35.8 | 29.4
Solution: We use Snell’s law: nair sin !air = nice sin !ice . The angle in diamond can be ignored.
6. Yellow light has a wavelength of 589 nm. What is the frequency and speed of the light in a
medium with an index of refraction 2.14 | 1.89 | 1.55 (frequency, speed)? (in Hz and m/s)
Answer: 5.09 × 1014, 1.40 ×108 | 5.09 × 1014, 1.59 ×108 | 5.09 × 1014, 1.94 ×108
Solution: The frequency is unchanged and given by f = c / ! = 3 #108 / 589 #10"9 = 5.09 #1014
Hz. The speed is cn = c / n , which depends on n.
7. A solenoid consists of a tube of diameter 24 cm and length 60 cm that has 2,500 turns of wire
wrapped around it. A current of 45 A flows in the wire. What is the inductance of the solenoid?
Answer: 0.59 H | 1.33 H | 2.61 H
Solution: The inductance is µ0 AN 2 / L , where N is the number of turns, L is the coil and A is the
coil area.
8. Refer to the previous problem. What is the energy contained in the magnetic field of the solenoid? (You can obtain this without solving the previous problem.)
Answer: 600 J | 1350 J | 2640 J
Solution: The energy U can be found from the B field. B = µ0 ni and U = B 2 AL / 2 µ0 , where n
= N / L.
9. A 60 Ω | 75 Ω | 90 Ω resistor and an inductor L are hooked in series to a 24 V battery and a
switch that is initially open, preventing current from flowing. A voltage of 17.0 V is measured
across the inductor 12 | 14 | 16 msec after the switch is closed. What is L?
Answer: 2.1H | 3.0 H | 4.2 H
Solution: The voltage across L is VL = V ! iR , where i = (V / R ) 1 ! e!tR / L . Thus VL = Ve!tR / L ,
(
and we can solve for L.
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PHY2054 Spring 2009
10. A 50 Ω resistor, a 150 mH inductor and a capacitor C are connected in series to an AC emf
source with Vrms = 36 V and variable frequency. It is found that when the source frequency is set
to 180 Hz | 210 Hz | 240 Hz the power delivered to the circuit is maximized. What is C?
Answer: 5.21 µF | 3.83 µF | 2.93 µF
Solution: The power is maximized at resonance ! = 1/ LC , which allows us to solve for C.
11. Refer to the previous problem. At resonance, what is the rms voltage across the capacitor?
Answer: 122 V | 143 V | 163 V
Solution: The voltage across C is the same as the voltage across L at resonance. The voltage
across C and L is given by VC = VL = (Vrms / R ) X L = (Vrms / R )!0 L , where !0 = 2" f 0 .
12. A type of glass has an index of refraction that varies from 1.62 at the violet end of the visible
spectrum to 1.52 | 1.49 | 1.46 at the red end of the spectrum. If white light of all visible wavelengths shines on a flat surface of the glass, what is the maximum angular dispersion of the refracted light?
Answer: 3.0° | 4.0° | 5.1°
Solution: The maximum dispersion is the maximum difference in refracted angle, which occurs
when the incident angle is 90°. Thus the dispersion is !disp = sin "1 (1/ nviolet ) " sin "1 (1/ nred ) .
13. Microwaves of frequency 5.50 GHz travel through a particular material of length 25.4 | 32.7 |
39.8 m at a speed 9% slower than the speed of light in vacuum. Approximately how many complete wavelengths fit inside the material?
Answer: 515 | 665 | 810
Solution: The wavelength inside the material is !n = 0.91c / f . The number of wavelengths that
fit inside the material is N ! = L / !n . The numbers shown are actually off by about 1% from the
true answers.
14. A single loop of wire with a radius of 7.5 | 5.0 | 5.0 cm rotates about a diameter in a uniform
magnetic field of 1.6 | 1.6 | 2.5 T. To produce a maximum emf of 1.0 V, it should rotate at:
Answer: 35 | 80 | 51 rad/s
Solution: The maximum emf is given by ! AB , where ω is the angular velocity, A = ! r 2 is the
area and B is the magnetic field strength.
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PHY2054 Spring 2009
15. A 0.20-cm | 0.10-cm | 0.20-cm radius cylinder, 3.0 | 4.0 | 5.0 cm long, is wrapped with wire
to form an inductor. At the instant the magnetic field in the interior is 5.0 | 10.0 | 5.0 mT the energy stored in the field is about:
Answer: 3.8 × 10−6 J | 5.0 × 10−6 J | 6.3 × 10−6 J
Solution: The energy stored in the field is B 2 AL / 2 µ0 , where A is the area of the solenoid, B is
the magnetic field strength and L is the length.
16. A solar cell has a light-gathering area of 10 | 10 | 20 cm2 and produces 0.20 | 0.40 | 0.20 A at
0.80 | 1.10 | 0.80 V (dc) when illuminated with sunlight of intensity 1000 W/m2. What is the efficiency of the solar cell?
Answer: 16% | 44% | 8%
Solution: The total power incident on the panel is IA, where I is the intensity and A is the area.
The output power is Vi, where i is the current and V is the voltage. The efficiency is given by the
ratio of these or ! = Vi / IA .
17. A laser beam of wavelength 632.8 | 595 | 480 nm has intensity 13.3 | 19 | 25 W/m2. The amplitudes of the associated electric and magnetic fields, in V/m and µT, respectively are:
Answer: 100, 0.33 | 120, 0.40 | 137, 0.46
2
2
/ 2 µ0 = Emax
/ 2cµ0 . Note Emax = cBmax .
Solution: The intensity is given by I = cBmax
18. A microwave oven heats food by using electromagnetic waves that form a standing wave in
the oven. When 2.5 GHz waves are used this way, it is found, in a sample of food, that there are
burn marks spaced 5.0 | 4.8 | 5.5 cm apart. What is the index of refraction of the food sample for
these electromagnetic waves?
Answer: 1.20 | 1.25 | 1.09
Solution: This was taken from a chapter problem. The wavelength in air is ! = c / f = 12 cm.
The power is maximum twice within each cycle, so the difference between the burn marks (L) is
half the wavelength in the material. Thus L = !n / 2 = ! / 2n and we can determine n.
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PHY2054 Spring 2009
19. The front end or antenna input of radio and television receivers is a resonant LC circuit (similar to that discussed in Chapter 21. Tuning to different channels (frequencies) is accomplished by
varying L or C until the circuit resonates at the station's frequency. If a particular circuit includes
a 1.0 microhenry inductor, what capacitance (in picofarads) is required to tune it to an FM station
that broadcasts on a wavelength of 3 | 4 | 5 meters?
Answer: 2.5 | 4.5 | 7.0
Solution: The resonant angular frequency ω0 is !0 = 1/ LC and is also equal to 2! f , where
f = c / ! . Thus we can determine C.
20. An RL circuit with resistance 100 Ω | 120 Ω | 150 Ωis connected in series to a battery with
EMF 12V. If the current builds up to 1/3 of its final steady-state value in 6.0 sec, what is the inductance L in Henrys?
Answer: 1480 | 1330 | 1290
Solution: The current in an RL circuit is i = (V / R ) 1 ! e!tR / L . The condition that 1/3 of the
(
)
maximum power is reached means that we have to solve e!tR / L = 2 / 3 for L.
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