This test is 90 minutes long and has a total of 100 points.
[15 points] 1. Multiple choice questions. Each question is worth 3 points. Partial credit may be given for an incorrect answer if work is shown.
(i). A uniform magnetic field B = 0.50 T is directed perpendicular to the plane of a circular loop of wire of radius 0.25 m. The magnetic flux through the loop is:
[ ] zero
[ ] 0.031 T m2
[ x ] 0.098 T m2
[ ] 2.55 T m2
There is no induced EMF, but there is certainly some magnetic flux!
(ii). A generator of AC electricity contains a coil with 25 turns (i.e., N = 25) in a uniform magnetic field of strength B = 0.33 T. The coil has an area of 0.50 m2, and is spinning at a frequency f = 60 Hz. The axis of rotation of the coil is perpendicular to the magnetic field (this is the standard arrangement). What is the peak voltage generated by this generator?
[ ] 248 V
[ ] 350 V
[ x ] 1555 V
[ ] 2200 V
(iii) The average distance between the Earth and the Sun is 1.49 x 1011 m. How much time, in minutes, does it take for light to travel from the Sun to the Earth?
[ ] zero
[ ] 2.01 x 10-3 minutes
[ ] 0.138 minutes
[ x ] 8.28 minutes
[ ] 497 minutes
(iv). An astronomer observes that electromagnetic waves emitted by oxygen atoms in a distant galaxy have a frequency of 5.710 x 1014 Hz. In a lab on Earth, oxygen atoms emit waves with a frequency of 5.841 x 1014 Hz. Determine the relative velocity of the galaxy with respect to the Earth.
[ x ] 6.73 x 106 m/s, away from the Earth.
[ ] 6.73 x 106 m/s, toward the Earth.
[ ] 4.37 x 106 m/s, away from the Earth.
[ ] 4.37 x 106 m/s, toward from the Earth.
[ ] 3.36 x 106 m/s, away from the Earth.
[ ] 3.36 x 106 m/s, toward from the Earth.
(v). Unpolarized light with an intensity of 400 W / m2 is incident on a polarizer; the polarization axis of the polarizer is at 20° to the vertical. What is the intensity of the transmitted light?
The intensity of the transmitted light is 200 W / m2. The incident light is UNpolarized, so the angle of the polarizer is irrelevant.
[20 points] 2. Induction.
A conducting rod of length L = 0.10 m lies on a pair of conducting rails in a region where a uniform magnetic field of 2.0 T is directed into the plane of the paper, as shown. The battery has an emf of 6 V, and the resistor has a resistance of R = 0.3 ohms. (Everything else has negligible resistance.)
[4 points] (a) The rod is initially clamped in place so that it can not move. What is the current in the rod?
The current is 20 A. If the rod does not move we have a simple circuit, and we can apply V = IR.
[4 points] (b) The rod is now unclamped so it is free to move without friction on the rails. In which direction will it move? Justify your answer.
The rod moves to the right. With the battery connected in the circuit as shown, current flows clockwise around the circuit. This means it is going down through the rod. If the current is down and the magnetic field is into the page, the right-hand rule tells us there'll be a force to the right. This force causes the rod to accelerate to the right.
[6 points] (c) As the rod moves, a motional emf is developed in it. Does this motional emf cause the current in the loop to increase, decrease or does the current stay the same? Justify your answer.
The current in the rod will decrease. The motional EMF developed in the rod will oppose the battery. The battery wants current to flow clockwise, the motional EMF wants the current to flow counter-clockwise, and this reduces the current through the rod. In fact, the current will eventually drop to zero. (This is what part (d) is about.)
[6 points] (d) The rod is eventually observed to be moving with a constant velocity. What is this velocity? State both the magnitude and the direction. [Hint: what is the net force on the rod?]
If the rod moves with a constant velocity, there must be no net force on the rod. The only way there can be no force is if there is no current through the rod - that's because the motional EMF has increased until it exactly cancels the EMF of the battery. So, the motional EMF is 6 volts. Setting this equal to vBL and solving for v gives a speed of 30 m/s. The rod moved right to start with, and it must still be going right.
[20 points] 3. AC Circuits
A light bulb has a resistance of 240 ohms. It is connected to a standard wall socket (in North America a wall socket has Vrms = 110 V, f = 60.0 Hz).
[3 points] (a) Determine the rms current in the bulb.
The rms current is 110 / 240 = 0.458 A.
[3 points] (b) How much power is dissipated in the bulb?
One way to get this is P = V2 / R = 50.4 W.
[6 points] (c) Determine the value of the inductance, L.
The key here is that the current stays the same. If the current is the same and the voltage is the same, then the impedance must be 240 ohms. The resistor is 240 ohms by itself, so the capacitive reactance has to be equal to the inductive reactance (in other words, the circuit is at resonance).
Setting XL = XC and solving for L gives L = 0.00704 H.
[4 points] (d) What is the rms current in the RLC circuit?
Same as part (a), 0.458 A.
[4 points] (e) It is possible to connect the bulb to a 220 V wall socket (the European system) using only a transformer so that the power dissipated in the bulb is the same as that in part (a). Explain in detail how you would do this, suggesting possible values for the number of turns in the primary and secondary.
If the 220 V is on the primary side of the circuit, the secondary side is connected to the light bulb. The light bulb must have 110 V, so that's the secondary voltage. To get 110 V from 220 V, the transformer must have a 2:1 ratio of turns. If there were 200 turns in the primary and 100 turns in the secondary, for instance, that would do it.
[10 points] 4. Parallel rays.
Three parallel rays are incident on a converging lens; the lens has a focal length of 3.0 cm. In each of the two cases below, draw the path of each light ray and state the x and y coordinates of the point where the rays converge.
Assume the lens lies at the origin of an x-y coordinate system, and the grid has a spacing of 1 cm by 1 cm.
[5 points] (a) The rays are parallel to the principal axis, and shine on the upper half of the lens.
The rays converge at: x = 3 cm y = 0 cm
The lens sends all rays that are parallel to the principal axis through the focal point, which is at (3,0) in this coordinate system.
[5 points] (b) The rays are parallel to the principal axis; a plane mirror is placed at x = 2 cm, parallel to the y-axis.
The rays converge at: x = 1 cm y = 0 cm
The lens still sends the rays off in the direction of the focal point, at (3,0). They don't get there, though, because the mirror is in the way. At x = 2 cm, the rays would have traveled another 1 cm before converging if the mirror was not there; because they reflect off the plane mirror at the same angle, they travel 1 cm back before converging. Another way to look at it is that the focal point is reflected in the mirror; instead of being 1 cm behind it, it's 1 cm in front of it.
[15 points] 5. Refraction
A light source lies at the bottom of a pool of water. It emits a beam of light that strikes the interface between the water and an unknown medium at 45°. The index of refraction of water is n = 1.33.
[5 points] (a) If the refracted beam passes through point A (x = 1, y = 2), what is the index of refraction of the unknown medium?
The first step is to figure out the angle of refraction. The tangent of the angle is 1/2, so the angle is 26.57°. Putting this in Snell's law gives:
1.33 sin45° = n2 sin26.57°
which gives n2 = 2.10
[5 points] (b) If, instead, the refracted beam passes through point B (x = 2, y = 2), what is the speed of light in the unknown medium?
If the beam goes through point B, then it's direction does not change when it crosses the interface. The index of refraction of the medium must then be 1.33, the same as water. Solving for the speed of light gives:
n = c / v, so v = c / n = 3.00 x 108 / 1.33 = 2.26 x 108 m/s
[5 points] (c) What would the index of refraction of the unknown medium have to be for all the light to be totally internally reflected so it passed through point C? Is this possible? Justify your answer.
For the light to be totally internally reflected, the index of refraction of the unknown medium must be less than that of water. The maximum possible value of the index of refraction is obtained from the critical-angle equation. Here the critical angle is 45°, the angle of incidence.
sin45° = n2 / n1
so n2 = 1.33 sin45° = 0.94
The index of refraction of the unknown medium must be less than 0.94 for total internal reflection to occur. This is NOT possible! The index of refraction of a medium must be at least 1.00; a value lower than 1.00 implies that the speed of light in the medium is greater than the speed of light in vacuum, which is not possible.
[20 points] 6. A lens
A certain lens produces a real, inverted image of a particular object. The image is 30 cm away from the object, and is exactly half as large as the object.
[3 points] (a) What kind of lens produces this image?
Because the image is real and inverted, the lens must be a converging lens.
[6 points] (b) How far from the object is the lens placed? Hint : use the magnification equation, m = -di / do, and one other condition to determine di and do, or look at one of the principal rays.
Method one: the magnification is m = -0.5, so di / do = 0.5. The image is real, so the lens must be between the object and the image; this means that di + do = 30 cm. Combining these two equations gives 20 cm for the object distance, and 10 cm for the image distance.
Method two: one of the principal rays is a straight line that joins the tip of the image and the tip of the object. This is the ray that passes through the center of the lens. If you draw this ray, it crosses the principal axis 20 cm from the object. The lens must be there, so the object distance is 20 cm.
[5 points] (c) What is the focal length of the lens?
Knowing the object and image distances allows the focal length to be solved for, from the equation:
1 / di + 1 / do = 1 / f.
This gives f = 6.67 cm.
[6 points] (d) Draw a ray diagram on the diagram above to show how the image is produced by the lens. (Hint : draw in the lens first.)
