Mutasem Jarwan
Grade 10 ADV Physics
EOT 3 Sample Questions
2023/2024 Exam coverage
indicate the direction of EMF and induced current
Induced Electromotive Force (EMF): source of potential difference that moves charges, it is produced by a changing of magnetic field. e.g., generator.
4. Use unit substitution to show that the units of BLv are volts.
1. Which dimensional analysis is correct for the calculation of EMF?
A. (N·A·m)(J)
B. (N/A·m)(m)(m/s)
C. J·C
D. (N·m·A/s)(1/m)(m/s)
9. Electric Generator Explain how an electric generator works.
An EMF is induced in the armature of a generator as it is turned—by a mechanical force—in a magnetic field. When the generator is in a circuit, the EMF induces a current. As the armature rotates through 180°, the induced EMF—and current—reverse direction.
10. Generator Could you make a generator by mounting permanent magnets on a rotating shaft and keeping the coil stationary? Explain.
Yes; only relative motion between the coil and magnetic field is important. Note, this generator would not have much power as the relative velocities of the magnets and coil will be very small.
11. Bike Generator A small generator on your bike lights the bike’s headlight. What is the source of the energy for the bulb when you ride along a flat road?
12. Microphone Consider the microphone shown in Figure 3. What happens when the diaphragm is pushed in?
The rider provides the mechanical energy that turns the generator’s armature.
A current is induced in the coil.
Eddy Current: �current generated in any piece of metal (or sheet) moving through a magnetic field.��Eddy current is an induced current produces induced magnetic field �that opposes the motion of metal in either direction.
To reduce eddy currents circulation the metal constructed as thin layers.
Eddy currents are used to slow �movement of metal parts. �e.g., the braking system of some �trains and roller coasters.
e.g., stop swinging of laboratory �balance.
v
F
B
I
EMF = BLv → I
F = ILB → B
motion produces current
current produces field
Lenz’s Law:�magnetic field produced by induced current opposes the change of original field that produces induced current.
Binduced opposite to ΔBoriginal
FB opposite to motion
Faraday’s experiment
Oersted's experiment
Lenz’s law depends on the conservation of energy principle.
Faraday’s law calculate the magnitude of induced EMF, then induced current.
Lenz’s law explains the direction of induced current.
ΔB → EMF → Ii → Bi → FB
induced quantities
5. A generator develops a maximum potential difference of 170 V.
Practice problems (page 159)
6. The RMS potential difference of an AC household outlet is 117 V.
Alternating currents and generators
c. What is the resistance of the lamp when it is working?
b. A 60 W lamp is placed across the generator with an Imax of 0.70. What is the effective current through the lamp?
a. What is the effective potential difference?
If the RMS current through the lamp is 5.5 A, what is the lamp’s maximum current?
What is the maximum potential difference across a lamp connected to the outlet?
7. If the average power used over time by an electric light is 75 W, what is the peak power?
Practice problems (page 159)
8. CHALLENGE An AC generator delivers a peak potential difference of 425 V.
b. The resistance is 5.0×102 Ω. What is the effective current?
a. What is the Veff in a circuit connected to the generator?
Alternating currents and generators
14. Output Potential Difference Explain why the output potential difference of an electric generator increases when the magnetic field is made stronger. What is another way to increase the output potential difference?
The magnitude of the induced EMF is directly related to the strength of the magnetic field. A greater potential difference is induced in the conductor(s) if the field strength is increased. Because EMF = BLv (sin θ), you can also increase output potential difference by increasing the length of the wire or the velocity of the wire.
15. Critical Thinking A student asks, “Why does AC dissipate power? The energy going into a lamp when the current is positive is removed when the current is negative. The net current is zero.” Explain why this reasoning is wrong.
Power is the rate at which energy is transferred. Power is the product of I and V. When I is positive, so is V, and therefore, P is positive. When I is negative, so is V; thus, P is positive. Energy is always transferred through the lamp.
1. You move a straight wire that is 0.5 m long at a speed of 20 m/s vertically through a 0.4 T magnetic field pointed in the horizontal direction.
Practice problems (page 155)
2. A straight wire that is 25 m long is mounted on an airplane flying at 125 m/s. The wire moves in a perpendicular direction through Earth’s magnetic field (B = 5.0×10-5 T). What EMF is induced in the wire?
Electromotive Force
b. The wire is part of a circuit with a total resistance of 6.0 Ω. What is the current?
a. What EMF is induced in the wire?
3. A straight wire segment in a circuit is 30.0 m long and moves at 2.0 m/s perpendicular to a magnetic field.
Practice problems (page 155)
4. CHALLENGE A horseshoe magnet is mounted so that the magnetic field lines are vertical. You pass a straight wire between the poles and pull it toward you. The current through the wire is from right to left. Which is the magnet’s north pole? Explain.
Electromotive Force
b. The total resistance of the circuit is 5.0 Ω. What is the current?
a. A 6.0 V EMF is induced. What is the magnetic field?
Using a right-hand rule, the north pole is at the bottom.
Electromagnetic Waves
43. Water has a dielectric constant of 1.77. What is the speed of light in water?
42. What is the speed of an electromagnetic wave traveling through air?
Use c = 299,792,458 m/s in your calculation.
Practice problems (page 181)
Calculate the speed of electromagnetic waves in different mediums of different dielectric constants.
Electromagnetic Waves
45. CHALLENGE A radio signal is transmitted from Earth’s surface to the Moon’s surface, 376,290 km away. What is the shortest time a reply can be expected?
Practice problems (page 181)
Calculate the speed of electromagnetic waves in different mediums of different dielectric constants.
The signal must reach the moon then come back to earth
Electromagnetic Waves
Practice problems (page 177)
Apply the wave equation to calculate the wavelength, frequency, or speed of electromagnetic waves.
Electromagnetic Waves
Practice problems (page 177)
Apply the wave equation to calculate the wavelength, frequency, or speed of electromagnetic waves.
49.Radio Signals Radio antennas normally have metal rod elements that are oriented horizontally. From this information, what can you deduce about the directions of the electric fields in radio signals?
50. Digital Signals What are the advantages of storing and transmitting sound, pictures, and data as digital signals?
Digital signals can be stored reliably in computer memory and sent over long distances.
They can send more information in the same amount of time as AM or FM and are less affected by noise.
They must also be horizontal.
51. Antenna Design Would an FM antenna designed to be most sensitive to stations near 88 MHz be shorter or longer than one designed to receive stations near 108 MHz? Explain your reasoning.
Longer; lower-frequency waves would have longer wavelengths, so they use a longer antenna.
Wire antennas When an antenna is one-half the length of the wave it is designed to detect, the potential difference across its terminals is largest and the antenna is most efficient. Therefore, an antenna designed to receive radio waves is longer than one designed to receive microwaves.
Everyday uses of transformers
Long-distance transmission of electrical energy is economical only if very high potential differences are used.
High potential differences reduce the current required in the transmission lines, keeping the wasteful energy transformations low.
As shown in Figure 16, step-up transformers are used at power sources, where they can develop potential differences up to 480,000 V. When the energy reaches homes, step-down transformers reduce the potential difference to 120 V. Game systems, printers, laptop computers, and rechargeable toys have transformers inside their casings or in blocks attached to their cords. These small transformers further reduce potential differences from wall outlets to the 3 V–26 V range.
1. Violet light falls on two slits separated by 1.90 x 10–5 m. A first-order bright band appears 13.2 mm from the central bright band on a screen 0.600 m from the slits. What is λ?
Practice problems (page 196)
2. Yellow-orange light from a sodium lamp of wavelength 596 nm is aimed at two slits that are separated by 1.90 x 10–5 m. What is the distance from the central band to the first-order yellow band if the screen is 0.600 m from the slits?
Interference
Apply the relation (𝜆=𝑥d/L) to calculate the wavelength or to find an unknown distance in a double-slit investigation given the other values.
3. In a double-slit investigation, physics students use a laser with λ = 632.8 nm. A student places the screen 1.000 m from the slits and finds the first-order bright band 65.5 mm from the central line. What is the slit separation?
Practice problems (page 196)
4. CHALLENGE Yellow-orange light with a wavelength of 596 nm passes through two slits that are separated by 2.25 x 10–5 m and makes an interference pattern on a screen. If the distance from the central line to the first-order yellow band is 2.00 x 10–2 m , how far is the screen from the slits?
Interference
Apply the relation (𝜆=𝑥d/L) to calculate the wavelength or to find an unknown distance in a double-slit investigation given the other values.
Interference
Show that the intensity of bright bands decreases as you go farther from the central band (double slit interference with monochromatic light).
Interference of Coherent Light:
Young, double slit experiment:
The overlapping light from the two slits fell on an observing screen.
The overlap created a pattern of bright and dark bands called interference fringes.
- Constructive interference produces a bright central band of the given color on the screen, as well as other bright bands of near-equal spacing and near-equal width on either side, as shown in Figure 3.
- The intensity of the bright bands decreases the farther the band is from the central band, as you can see.
- Between the bright bands are dark areas where destructive interference occurs.
- The positions of the constructive and destructive interference bands depend on the light's wavelength.
Thin film interference:
the phenomenon of a spectrum of colors resulting from constructive and destructive interference of light waves due to reflection from separate surfaces in a thin film.
Thin film interference:
the phenomenon of a spectrum of colors resulting from constructive and destructive interference of light waves due to reflection from separate surfaces in a thin film.
Electromotive Force
Induced EMF A straight wire is part of a circuit that has a resistance of 0.50 Ω. The wire is 0.20 m long and moves at a constant speed of 7.0 m/s perpendicular to a magnetic field of strength 8.0 x 10–2 T.
Example problem (page 154)
(a) What EMF is induced in the wire?
(b) What is the current through the wire?
(c) If a different metal were used for the wire, increasing the circuit’s resistance to 0.78 Ω, what would the new current be?
1. You move a straight wire that is 0.5 m long at a speed of 20 m/s vertically through a 0.4 T magnetic field pointed in the horizontal direction.
Practice problems (page 155)
2. A straight wire that is 25 m long is mounted on an airplane flying at 125 m/s. The wire moves in a perpendicular direction through Earth’s magnetic field (B = 5.0×10-5 T). What EMF is induced in the wire?
Electromotive Force
b. The wire is part of a circuit with a total resistance of 6.0 Ω. What is the current?
a. What EMF is induced in the wire?
3. A straight wire segment in a circuit is 30.0 m long and moves at 2.0 m/s perpendicular to a magnetic field.
Practice problems (page 155)
4. CHALLENGE A horseshoe magnet is mounted so that the magnetic field lines are vertical. You pass a straight wire between the poles and pull it toward you. The current through the wire is from right to left. Which is the magnet’s north pole? Explain.
Electromotive Force
b. The total resistance of the circuit is 5.0 Ω. What is the current?
a. A 6.0 V EMF is induced. What is the magnetic field?
Using a right-hand rule, the north pole is at the bottom.
STEP-UP TRANSFORMERS: A step-up transformer has a primary coil consisting of 200 turns and a secondary coil consisting of 3000 turns. The primary coil is supplied with an effective AC potential difference of 90.0 V.
Example problem (page 166)
b. The current in the secondary circuit is 2.0 A. What is the current in the primary circuit?
a. What is the potential difference in the secondary circuit?
Transformers
16. A step-down transformer has 7500 turns on its primary coil and 125 turns on its secondary coil. The potential difference across the primary circuit is 7.2 kV.
Practice problems (page 166)
c. Calculate input power
b. If the current in the secondary circuit is 36 A, what is the current in the primary circuit?
a. What is the potential difference across the secondary circuit?
Transformers
17. CHALLENGE A step-up transformer has 300 turns on its primary coil and 90,000 turns on its secondary coil. The potential difference of the generator to which the primary circuit is attached is 60 V. The transformer is 95 percent efficient.
Practice problems (page 166)
b. The current in the secondary circuit is 0.5 A. What current is in the primary circuit?
a. What is the potential difference across the secondary circuit?
Transformers
Transformers
Lenz’s Law
a) The direction of original magnetic field (from the magnet):
Determine the type of pole induced on the face of a coil and the direction of induced current in a coil when a coil and magnet are in relative motion.
2. Look at the figures and answer the following questions:
Extra Questions
b) The induced magnetic pole appears on the left side of coil is:
c) The direction of the induced current passes through galvanometer:
Right, field lines com out from north and enter the south.
North, the magnet is moving towards the coil. Original field increase, the coil will induce a field in the opposite direction to the original field.
From left to Right, using the right hand rile. (thump with the direction of the induced field , fingers show the direction of the induced current)
Lenz’s Law
Determine the type of pole induced on the face of a coil and the direction of induced current in a coil when a coil and magnet are in relative motion.
3. Look at the figures and answer the following questions:
Extra Questions
a) The direction of original magnetic field (from the magnet):
b) The induced magnetic pole appears on the left side of coil is:
c) The direction of the induced current passes through galvanometer:
Right, field lines com out from north and enter the south.
South, the magnet is moving away from the coil. Original field decrease, the coil will induce a field in the same direction as the original field.
From right to left, using the right hand rile. (thump with the direction of the induced field , fingers show the direction of the induced current)
Lenz’s Law
Determine the type of pole induced on the face of a coil and the direction of induced current in a coil when a coil and magnet are in relative motion.
4. Look at the figures and answer the following questions:
Extra Questions
a) The direction of original magnetic field (from the magnet):
b) The induced magnetic pole appears on the left side of coil is:
c) The direction of the induced current passes through galvanometer:
Left, field lines com out from north and enter the south.
South, the magnet is moving towards the coil. Original field increase, the coil will induce a field opposite to the original field.
From right to left, using the right hand rile. (thump with the direction of the induced field , fingers show the direction of the induced current)
Interference
Define coherent and incoherent light.
Differentiate between incoherent and coherent waves by giving examples like rain or water droplets falling on the surface of water.
Incoherent and Coherent Light:
Incoherent light: A light whose waves are not in phase, such as white light from a light bulb.
The effect of incoherence in waves can be seen in the example of heavy rain falling on still water. The surface of the water is choppy and does not have a regular pattern of waves, as shown in Figure 1. Because light waves have such a high frequency, incoherent light does not appear choppy to you. Instead, as light from an incoherent white light source illuminates an object, you see the combination of the incoherent light waves as an even, white light.
Coherent light: Light made up of waves of the same wavelength that are in phase with each other.
A regular wavefront, which is made of coherent light, can be created by a single point source, as shown in Figure 2. A regular wavefront also can be created by multiple point sources when all point sources are in phase. This type of coherent light is produced by a laser.
Interference
Recall the concepts of constructive and destructive interference and define interference fringes of light.
Interference:
The result of the superposition of tow or more waves.
Constructive interference, in which two waves (of the same wavelength) interact in such a way that they are aligned, leading to a new wave that is bigger than the original wave
Destructive interference, in which two waves (with the same amplitude) are shifted by exactly half a wavelength when they merge, they will cancel each other out.
Interference
Apply the relation (𝜆=𝑥d/L) to calculate the wavelength or to find an unknown distance in a double-slit investigation given the other values.
WAVELENGTH OF LIGHT A double-slit investigation is performed to measure the wavelength of red light. The slits are 0.0190 mm apart. A screen is placed 0.600 m away, and the first-order bright band is 21.1 mm from the central bright band. What is the wavelength of the red light?
Example problem (page 196)
1. Violet light falls on two slits separated by 1.90 x 10–5 m. A first-order bright band appears 13.2 mm from the central bright band on a screen 0.600 m from the slits. What is λ?
Practice problems (page 196)
2. Yellow-orange light from a sodium lamp of wavelength 596 nm is aimed at two slits that are separated by 1.90 x 10–5 m. What is the distance from the central band to the first-order yellow band if the screen is 0.600 m from the slits?
Interference
Apply the relation (𝜆=𝑥d/L) to calculate the wavelength or to find an unknown distance in a double-slit investigation given the other values.
3. In a double-slit investigation, physics students use a laser with λ = 632.8 nm. A student places the screen 1.000 m from the slits and finds the first-order bright band 65.5 mm from the central line. What is the slit separation?
Practice problems (page 196)
4. CHALLENGE Yellow-orange light with a wavelength of 596 nm passes through two slits that are separated by 2.25 x 10–5 m and makes an interference pattern on a screen. If the distance from the central line to the first-order yellow band is 2.00 x 10–2 m , how far is the screen from the slits?
Interference
Apply the relation (𝜆=𝑥d/L) to calculate the wavelength or to find an unknown distance in a double-slit investigation given the other values.
Interference
Show that the intensity of bright bands decreases as you go farther from the central band (double slit interference with monochromatic light).
Interference of Coherent Light:
Young, double slit experiment:
The overlapping light from the two slits fell on an observing screen.
The overlap created a pattern of bright and dark bands called interference fringes.
- Constructive interference produces a bright central band of the given color on the screen, as well as other bright bands of near-equal spacing and near-equal width on either side, as shown in Figure 3.
- The intensity of the bright bands decreases the farther the band is from the central band, as you can see.
- Between the bright bands are dark areas where destructive interference occurs.
- The positions of the constructive and destructive interference bands depend on the light's wavelength.
Interference
Explain the formation of a colored spectra when white light is used in a double-slit investigation.
Interference of Coherent Light:
- Because the positions of the other bright bands of constructive interference depend on wavelength, each color’s band is at a different position, resulting in spectra of color.
What happens when a white light is used in a double-slit investigation?
- The interference causes the appearance of colored spectra.
- The various bands of color from the visible spectrum overlap on the screen. All these colors have constructive interference, and the central band is white.
Thin-film interference
Solve problems on interference of light.
OIL AND WATER You observe colored rings on a puddle and conclude that there must be an oil slick on the water. You look directly down at the puddle and see a yellow-green (λ = 555 nm) region. If the refractive index of oil is 1.45 and that of water is 1.33, what is the minimum thickness of oil that could cause this color?
Example problem (page 199)
Because you want the minimum thickness, m = 0.
5. In the situation in Example Problem 2, what would be the thinnest film that would create a reflected red (λ = 635 nm) band?
Practice problems (page 199)
6. A glass lens has a nonreflective coating of magnesium fluoride placed on it. How thick should the nonreflective layer be to keep yellow-green light with a wavelength of 555 nm from being reflected? See the sketch in Figure 9.
Thin-film interference
Solve problems on interference of light.
Because you want the minimum thickness, m = 0.
7. You can observe thin-film interference by dipping a bubble wand into some bubble solution and holding the wand in the air. What is the thickness of the thinnest soap film at which you would see a black stripe if the light illuminating the film has a wavelength of 521 nm? Use n = 1.33 for the bubble solution.
Practice problems (page 199)
8. What is the thinnest soap film (n = 1.33) for which light of wavelength 521 nm will constructively interfere with itself?
Thin-film interference
Solve problems on interference of light.
Because you want the minimum thickness, m = 1.
For Constructive interference
9. CHALLENGE A silicon solar cell has a nonreflective coating placed on it. If a film of silicon monoxide, n = 1.45, is placed on the silicon, n = 3.5, how thick should the layer be to keep yellow-green light (λ = 555 nm) from being reflected?
Practice problems (page 199)
Thin-film interference
Solve problems on interference of light.
Because you want the minimum thickness, m = 0.