PHY 2054 2nd Exam (SF 18–21) SOLUTIONS 6 July 2010
Each solution is pasted in, following the list of possible answers. Answer (1) is the correct one.
As you can see, the diagrams didn't make it through the file translations; some Greek letters and sub/superscripts may suffer similar casualties — If you don't have your copy of the exam handy, you may have to use your imagination!
1. Immediately after switch S in the circuit shown is closed, the current through the battery shown is:
(1) Vo/(R1 + R2) (2) Vo/R1 (3) Vo/R2 (4) 0 (5) Vo/(R1 + R2)/(R1R2)
1. Initially, the back emf of the inductor L blocks any current through L.
So I = V/(R1 + R2)
2. In the diagrams, all light bulbs are identical and all emf devices are identical. In which circuit (I, II, III, IV, V) will the bulbs be dimmest?
(1) IV (2) I (3) II (4) III (5) V
2. Circuit (IV) has the lowest potential (one battery; the parallel battery doesn't raise the voltage) across the highest resistance. So these bulbs are dimmest.
3. In the circuit shown, the capacitor is initially uncharged and V = 9.0 volts. At time t = 0, switch S is closed. If τ denotes the time constant, the approximate current through the 3Ω resistor when t = τ/100 is:
(1) 1A (2) 1/2A (3) 3/4A (4) 3/8A (5) 3/2A
3. t = RC = (3 + 6)W (6 mF) = 54 ms . Thus t/100 = 540 ns, So i = E/R (1 - e^(-100) = 9v/9V [1 - 0] = 1A
4. The Žgure shows the motion of electrons in a wire which is near the N pole of a magnet. The wire will be pushed:
(1) upwards (out of the page)
(2) toward the magnet
(3) away from the magnet
(4) downwards (into the page)
(5) along its length
4. By the right hand rule [RHR], the force on the wire is upward.
5. A square loop of wire lies in the plane of the page and carries a current I as shown. There is a uniform magnetic Želd B parallel to the side MK as indicated. The loop will tend to rotate:
(1) about PQ with KL coming out of the page
(2) about PQ with KL going into the page
(3) about RS with MK coming out of the page
(4) about RS with MK going into the page
(5) about an axis perpendicular to the page
5. Again, by the RHR (curled fingers) the magnetic moment of the loop is down into the page.
The torque is in a direction so as to bring the magnetic moment into alignment with the external field B.
So the loop rotates about axis PQ with the top edge KL coming up out of the page.
6. Two long straight wires pierce the plane of the paper at vertices of an equilateral triangle as shown at right. They each carry 2 A, out of the page. The magnetic Želd at the third vertex (P) has magnitude (in microTeslas):
(1) 17 (2) 10 (3) 20 (4) 50 (5) 8.7
6. Each current produces B = m0 I/(2 pr). Here, r = 4 cm. The field lines are circles and at P, the fields are *tangent* to the circles and perpendicular to the sides of the triangle. Thus the resultant B is the vector sum of the fields of the two currents :
B = 2 m0 I/(2 pr) cos 30” = 2 (4p x10^-7) 2A/(2p x 4 x 10^-2) (0.866) = 1.73 x 10^-5 = 17mT
7. An 8.0-mH inductor and a 2.0-Ω resistor are wired in series to a 20-V ideal battery. A switch in the circuit is closed at time 0, at which time the current is 0. A long time after the switch is thrown the potential diﬀerences across the inductor and resistor are, respectively:
(1) 0, 20 V
(2) 20 V, 0
(3) 10 V, 10 V
(4) 16 V, 4 V
(5) unknown since the rate of change of current is not given
7. After a long time (many time constants) the inductor current is just the current through the resistor = E/R . The battery's potential difference appears entirely across R . So 0V, 20V.
8. An LC circuit has an oscillation frequency of 105 Hz. If C=0.1µF, then L must be approximately:
(1) 25 µH (2) 10 mH (3) 1 mH (4) 2.5 µH (5) 1 pH
8. At resonance , f = 1/ [2p Sqrt(LC)] . So L = 1/[(2p f)^2 C] = 1/ (39500) = 2.5 e(-5) = 25m H
9. The primary of a 1:3 step-up transformer is connected to an AC source and the secondary is connected to a resistor R. The power dissipated by R in this situation is P. If R is connected directly to the source it will dissipate a power of:
(1) P/9 (2) P/3 (3) P (4) 3P (5) 9P
9. V1/V2 = I1/I2 = P1/P2 = N1/N2 = 1/3 . So I1 = 3I2 ==> I2 = I1/3 .
Secondary power P2 = (I2)^2 R = (I1/3)^2 R = (I1^2)R/9 = P2/9
10. A long straight wire is in the plane of a rectangular conducting loop. The straight wire carries a constant current i, as shown. If the wire is moved toward the rectangle at constant speed, the induced current in the rectangle is:
(4) clockwise in the left side and counterclockwise in the right side
(5) counterclockwise in the left side and clockwise in the right side
10. The B-field of the long wire is a series of circles around the wire, going down into the page and strongest next to the wire. Thus the flux through the loop is increasing as the wire moves toward it. By Lenz's Law, the induced current in the loop must produce B *opposing* the change, so loop current is ccw.
11. The circuit shown is immersed in a uniform magnetic Želd that is into the page and is decreasing in magnitude at the rate 150 T/s. The current in the circuit (in amperes) is:
(1) 0.18 (2) 0.22 (3) 0.40 (4) 0.62 (5) none of these
11. There are two currents: (a) battery current = E/R = 4V/10W = 0.4 V, ccw '
(b) induced current = (induced emf) / R ; emf = (area) x dB/dt = (0.12m)^2 x 150 = 2.16 V .
Then I = 2.16/10 = 0.216A. B is decreasing into the page so I must produce B in same direction cw.
So the two currents oppose each other: I = 0.4 - 0.216 = 0.18A
12. When the switch S in the circuit shown is closed, the time constant for the growth of current in R2 is:
(1) L/R2 (2) L/R1 (3) L/(R1 +R2) (4) L(R1 +R2)/(R1R2) (5) (L/R1 + L/R2)/2
12. The current through L is determined by R2, so TC = L/R2.
13. A large jetliner with a wingspan of 40 m ßies horizontally and due north at a speed of 300 m/s in a region where the magnetic Želd of the earth is 60µT directed 50◦ below the horizontal. What is the magnitude of the induced emf between the ends of the wing, in millivolts?
(1) 550 (2) 250 (3) 350 (4) 750 (5) none of these
13. For this example, E = Bwv, where w = wingspan and B is the perpendicular B- field component.
Thus E = 60 e-6 T x 40 m (sin 50” ) x 300 m/s = 0.55 e-6 = 550 e-3 = 550mV
14. A ßat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic Želd that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s. If the coil has a total resistance of 0.40Ω, what is the magnitude of the induced current, in milliamperes?
(1) 500 (2) 70 (3) 140 (4) 800 (5) none of these
14. E = -N dF /dt = N (area) dB/dt = 20 x 50e-4 m^2 x 2T/s = 0.2V
I = 0.2/0.4 = 0.5A = 500 mA
15. A bar magnet is falling through a loop of wire with constant velocity. The south pole enters Žrst. As the magnet leaves the loop, the induced current in the loop (as viewed from above):
(1) is counterclockwise (2) is clockwise (3) is zero (4) is along the length of the magnet (5) none of these
15. As the magnet leaves the loop, we have an N-pole (B upward) moving *away* from the loop and thus B is decreasing .
Thus by Lenz's Law, Induced I must produce B in same direction (up) so by RHR I is CCW.
16. An incredible amount of electrical energy passes down a funnel of a large tornado every second. Measurements taken near the earthÕs surface in Oklahoma at a distance of 9.00 km from a large tornado showed an almost constant horizontal magnetic Želd of 1.50 x 10-8T associated with the tornado. What was the average current going down the funnel?
(1) 675 A (2) 450 A (3) 950 A (4) 1500 A (5) none of these
16. A horizontal field must be due to the vertical current (like that of a long straight wire).
Thus I = 2p rB/( 4p e-7) = = (rB/2) e7 = 675 A
17. A proton is released such that its initial velocity is from right to left across the page. The protonÕs path, however, is deßected toward the bottom edge of the page due to the presence of a uniform magnetic Želd. What is the direction of this Želd?
(1) down, into the page
(2) up, out of the page
(3) from the bottom edge of the page, toward the top edge
(4) from right to left across the page
(5) none of these
17. by the RHR, B must be down, into the page
18. If I =2.0 mA and the potential diﬀerence, VA - VB = +30V in the circuit segment shown, determine the charge and polarity of the capacitor.
(1) 1.5 mC, left plate is positive
(2) 1.5 mC, right plate is positive
(3) 0.50 mC, left plate is positive
(4) 0.50 mC, right plate is positive
(5) none of these
18. The voltage across R is IR = 20V, same polarity orientation as the battery.
Moving from A to B, we have VC + 40V + 20V = +30V ==> VC= -30V
Q = CV = 50m F x -30V = -1500 mC = -1.5 mC The minus sign means that C's voltage (and hence charge) is opposite in polarity
to that of R and the battery, so left plate is positive.
19. Which device(s): resistors (R), capacitors (C), inductors (L), dissipate no energy, on average, in alternating current circuits?
(1) C and L (2) R, C,and L (3) R only (4) C only (5) L only
19. Only resistors dissipate energy in ac circuits
20. Commercial electric lanterns contain a heavy-duty battery (usually 6-volt) in series with an incandescent bulb and a switch. Here are in-use electrical data from a typical lantern:
Battery terminal voltage (switch OFF): 5.6 V
Battery terminal voltage (switch ON): 4.0 V
Circuit current (switch ON): 0.4 A
This battery's internal resistance in ohms is:
(5) none of these
20. The terminal voltage decreased (5.6V - 4V) = 1.6V with the circuit closed. This corresponds to the voltage drop across the internal resistance,
so r = V/I = 1.6/0.4 = 4 ohms