Zeena Husain, Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page nd all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points At some time after the switch S is closed, there is a current I owing through the resistor and inductor which is increasing in time dI > 0 (see the gure below). At this time dt there is an amount of charge of magnitude q on each plate of the capacitor.
1
dI the emf induced by the inductor is L . dt Summing the potential dierences around the loop gives E IRL dI q/C = 0 . dt
002 (part 2 of 2) 10 points
4.18 M 18 V
5.48 mH 7.07 F S
R E
I S
L C In the circuit shown above, what is the charge on the capacitor after the switch has been closed for a long time? Correct answer: 0.00012726 C. Explanation: Let : E = 18 V , R = 4.18 M , L = 5.48 mH , C = 7.07 F .
Which of the following equations is correct? 1. none of these 2. E + I R L 3. E I R L 4. E I R + L 5. E + I R L 6. E I R L 7. E + I R + L 8. E + I R + L 9. E I R + L dI dt dI dt dI dt dI dt dI dt dI dt dI dt dI dt + + + + q C q C q C q C q C q C q C q C =0 = 0 correct =0 =0 =0 =0 =0 =0 003 (part 1 of 1) 10 points Assume: The induced emf for the closed loop octagonal CXDY C is E . A solenoid (with magnetic eld B ) produces a steadily increasing uniform magnetic
and
After a long time t the capacitor will reach its maximum charge qmax . When this occurs the current will drop to zero. Setting I = 0 in the loop equation from Part 1 gives E or qmax = C E = (7.07 F) (1 106 F/F) (18 V) = 0.00012726 C , qmax = 0, C
Explanation: Apply Kirchos loop rule to the above cirdI > 0; therefore cuit. Here we are told that dt
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban ux through its circular cross section. A octagonal circuit surrounds the solenoid as shown in the gure. The wires connecting in the circuit are ideal, having no resistance. The circuit consists of two identical light bulbs (labeled X and Y ) in series. A wire connects points C and D. The ratio of the solenoids area AL left of the wire CD and the solenoids AL area AR right of the wire CD is = 4. AR i2 D
2
Since
AL = 4, we can solve for AL and AR in AR terms of A. AL = 4A 5 AR = A . 5
Then we can compute the magnitude of the induced emf around the right and left loops. ER = A R dB A dB 1 = =E dt 5 dt 5 dB 4A dB 4 EL = A L = = E. dt 5 dt 5
i3
i1
B
B
The induced emf and the changing magnetic ux are related by
X
Y
AL
B
AR
B
E =
dB d = A . dt dt
C
The equations for the (right) loop CXDC and the (left) loop CDY C are respectively given by 4E E i2 R = 0 . 1. + i1 R = 0 and 5 5 E 4E + i1 R = 0 and + i2 R = 0 . 2. 5 5 E 4E 3. i1 R = 0 and i2 R = 0 . 5 5 4E E i1 R = 0 and + i2 R = 0 . 4. 5 5 4E E 5. i1 R = 0 and i2 R = 0 . 5 5 E 4E 6. i1 R = 0 and + i2 R = 0 . 5 5 4E E + i2 R = 0 . cor7. + i1 R = 0 and 5 5 rect 4E E + i1 R = 0 and i2 R = 0 . 5 5 Explanation: By denition, the areas of the left and right loops are related by 8. A = A L + AR .
Since the magnetic ux is increasing, the induced emf is in the clockwise direction and the direction of the current is counter-clockwise, as shown in the gure. From Kirchos laws, the loop equations for the right and left loops respectively are right loop : left loop : 1 E + i1 R = 0 5 4 E + i2 R = 0 5 (1) (2)
004 (part 1 of 1) 10 points In the gure shown, the north pole of the magnet is rst moved down toward the loop of wire, then is withdrawn upward.
N Counterclockwise
Clockwise
As viewed from above, the induced current in the loop is 1. rst clockwise, then counterclockwise. 2. for both cases counterclockwise with decreasing magnitude. 3. rst counterclockwise, then clockwise. correct
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban 4. for both cases counterclockwise with increasing magnitude. 5. for both cases clockwise with decreasing magnitude. 6. for both cases clockwise with increasing magnitude. Explanation: From Ohms law and Faradays law, the 1 d V = , current in magnitude is I = R R dt where is the magnetic ux through the loop. We know the sign of the rate of change of the magnetic ux is changed when the magnet is withdrawn upward, which, according to the equation the direction of the current is also changed. From Lenzs law, we know when the magnet is moved down toward the loop, the current in the loop is counterclockwise as viewed from above. 005 (part 1 of 2) 10 points To demonstrate the generation of electromagnetic waves due to oscillations of a current sheet in the yz -plane at x = 0, one rst considers the following situation. One turns on a steady current ow along the negative y -axis at t = 0.
3
2. B is along the negative y -axis, E is along the positive z -axis. 3. B is along the negative z -axis, E is along the negative y -axis. 4. B is along the negative y -axis, E is along the positive y -axis. 5. B is along the positive z -axis, E is along the positive y -axis. 6. B is along the positive y -axis, E is along the negative z -axis. 7. B is along the positive y -axis, E is along the positive z -axis. 8. B is along the negative y -axis, E is along the negative z -axis. 9. B is along the negative z -axis, E is along the positive z -axis. 10. B is along the negative z -axis, E is along the positive y -axis. correct
y 0 I
p z
x
The point P is along the negative xdirection. Let B and E be the magnetic and electric eld established at P soon after the switching on of the current, respectively. The directions of B and E are 1. B is along the positive z -axis, E is along the negative y -axis.
Explanation: Basic Concepts Electromagnetic waves. Since the current on the sheet is owing along the negative y -axis, the magnetic eld for negative values of x is directed along the negative z -axis.
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban
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B C y E B P z E P
A O D
B O x
About point P we may construct a rectangular window ABCDA in the xy -plane. As the front of the B -eld sweeps across the window to the left, it causes a counterclockwise induced emf, which is generated by the induced emf at the right side of the window from D to A. Since the side DA can be placed at any location between O and P , when the B -eld reaches P , there is a companion E eld stretched out from D to A. Notice that the electromagnetic wave is traveling along the negative x-axis, so E has to be directed along the positive y -axis (so that E B is directed along the correct direction). 006 (part 2 of 2) 10 points Now consider the current sheet which undergoes the sinusoidal oscillations with a denite frequency where the amplitude of the magnetic eld oscillations in the waves is Bmax . Consider a book of area A located at P parallel to the yz plane. Find the radiation force on the book, if it absorbs one third of the light.
2 1 Bmax A 1. 6 0 2 3 Bmax A 2. 5 0 2 5 Bmax A 3. 3 0
2 4 Bmax A 3 0 2 1 Bmax A 5. 3 0 2 5 Bmax A 6. correct 6 0 B2 A 7. max 0 2 2 Bmax A 8. 3 0 2 1 Bmax A 9. 2 0 B2 A 10. 5 max 0 Explanation: The book absorbs one third of the light and reects two thirds of it. The pressure on the book is then
4.
2 5 1 P = u + 2 u = u. 3 3 3 In the present context, the radiation force is constant, so u is the average electromagnetic 2 energy density. Since u = Bmax /(20 ), we have 2 5 Bmax A F = PA = . 6 0 007 (part 1 of 1) 10 points A point light source delivers a time-averaged power P . It radiates light isotropically. A piece of small at surface is placed at D, which is a distance r away. This piece has a cross 1 of the section Asurf . The surface reects 4 3 light and absorbs of the light. Assume the 4 light hitting the various parts of the surface is perpendicular to them.
r Point source
D
The time-averaged energy density hitting the surface is given by:
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban P 1. u = Asurf 2. u = 4 r 2 P 3. u = r2 P 4. u = r2 5. u = P c
5
Which expression gives the magnitude B (r3 ) at D of the magnetic eld in the region b < r = r3 < a? 1. B (r3 ) = 0 i r3 2 0 i (a 2 r 3 ) correct 2. B (r3 ) = 2 r 3 (a 2 b 2 ) 2 0 i (r3 b2 ) 3. B (r3 ) = 2 r 3 (a 2 b 2 ) 4. B (r3 ) = 0 5. B (r3 ) =
P 4 r2
6. u = Asurf P P 7. u = Asurf c P 8. u = 4 r 2 c P 9. u = c Asurf P 10. u = correct 4 c r2 Explanation: Basic Concepts EM Wave The time-averaged energy density at D is given by P I . u= = c 4 r2 c 008 (part 1 of 1) 10 points The gure below shows a coaxial cable of radii a, b, and c in which equal, uniformly distributed, but antiparallel currents i exist in the two conductors. a b iout c iin O F E D C r1 r2 r3 r4
0 i 2 r3 0 i r 3 6. B (r3 ) = 2 a2 0 i r 3 7. B (r3 ) = 2 b2 0 i (a 2 b 2 ) 8. B (r3 ) = 2 2 r 3 (r 3 b 2 ) 0 i r 3 9. B (r3 ) = 2 c2 2 0 i (a 2 + r 3 2 b 2 ) 10. B (r3 ) = 2 r 3 (a 2 b 2 ) Explanation: Amperes Law states that the line integral B .dl around any closed path equals 0 I , where I is the total steady current passing through any surface bounded by the closed path. Considering the symmetry of this problem, we choose circular a path, then Amperes Law is simplied to: B (2 r1 ) = 0 iin , where r1 is the radius of the circle and iin is the current enclosed. For Part 1, b < r = r3 < a, 0 Iin B= 2r (r 2 b 2 ) 0 i i (a 2 b 2 ) = 2r 2 r2 a 0 i a2 b 2 = 2r
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban = 0 i (a 2 r 2 ) . 2 r (a 2 b 2 ) Mutual Inductance: M= N2 21 N1 12 = I1 I2 E = M dI dt
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009 (part 1 of 1) 10 points A long solenoid has a coil inside of it made of ne wire coaxial with it. r R
d B The emf is given by E = N . The magdt netic ux is B = B A = (0 n I ) ( R2 ) , so E = R2 0 n N dI . dt
Inside coil has N turns Outside solenoid has n turns per meter Given a varying current I in the outer solenoid, what is the emf induced in the inner loop? dI dt dI 2. E = R 0 N dt dI 3. E = r 0 n N dt dI 4. E = R2 0 n N correct dt dI 5. E = r 0 N dt dI 6. E = r 0 n dt dI 7. E = R2 0 n dt dI 8. E = R 0 n N dt dI 9. E = r 2 0 n dt dI 10. E = r 2 0 n N dt Explanation: Basic Concepts: Magnetic Field of a Solenoid: B = 0 nI 1. E = R 0 n Faradays Law: E = d B dt
8T Calculate the applied force required to move the bar to the right at a constant speed of 6 m/s. Correct answer: 877.714 N. Explanation: Basic Concept: Motional emf E = B v.
m 1g
4m I
7
I
We are interested in the emf in the inner coil. We use the area the smaller area of the inner coil rather than the larger solenoid area. 010 (part 1 of 1) 10 points Given: Assume the bar and rails have negligible resistance and friction. In the arrangement shown in the gure, the resistor is 7 and a 8 T magnetic eld is directed into the paper. The separation between the rails is 4 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 6 m/s . 8T 6 m/s
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban Magnetic force on current F =I Ohms Law I= V . R B.
7
What is the magnitude of the magnetic eld at point P (at the center of the square loop) due to the current in only one of the sides of the wire? 1. B = 2. B = 3. B = 4. B = 5. B = 6. B = 7. B = 8. B = 9. B = 10. B = 2 0 I I 0 2 2 0 I 2 3 0 I 2 3 0 I 2 I 0 2 0 I 2 correct 2 0 I 2 0 I 22 0 I 2
Solution: The motional emf induced in the circuit is E =B v = (8 T) (4 m) (6 m/s) = 192 V . From Ohms law, the current owing through the resistor is I= E R 192 V = 7 = 27.4286 A .
Thus, the magnitude of the force exerted on the bar due to the magnetic eld is FB = I B = (27.4286 A)(4 m)(8 T) = 877.714 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = FB = 877.714 N 011 (part 1 of 2) 10 points A conductor in the shape of a square of edge length carries a counter-clockwise current I as shown in the gure below.
Explanation: Basic Concepts: The Biot-Savart law is dB = r 0 I ds . 2 4 r
Solution: Consider a thin, straight wire carring a constant current I along the x-axis with the y -axis pointing towards the center of the square, as in the following gure. y P r
a x
P r I ds x
O
I
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban Let us calculate the total magnetic eld at the point P located at a distance a from the wire. An element ds is at a distance r from P . The direction of the eld at P due to this element is out of the paper, since ds is r out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the eld at P . In fact, taking the origin at O and letting P be along the positive y axis, with k being a unit vector pointing out of the paper, we see that ds = k |ds | = k (dx sin ) . r r Substituting into Biot-Savart law gives dB = k dB , with 0 I dx sin . (1) 4 r2 In order to integrate this expression, we must relate the variables , x, and r. One approach is to express x and r in terms of . From the geometry in the gure and some simple dierentiation, we obtain the following relationship a r= = a csc . (2) sin a Since tan = from the right triangle in x the gure, dB = x = a cot and dx = a csc2 d . Substituting Eqs. 2 and 3 into Eq. 1 gives 0 I a sin d 2 csc2 4 a 0 I sin d . (4) = 4a Thus, we have reduced the expression to one involving only the variable . We can now obtain the total eld at P by integrating Eq. 4 over all elements subtending angles ranging from 1 to 2 as dened in the gure. This gives dB = 0 I 2 B= sin d 4 a 1 0 I (cos 1 cos 2 ) . = 4a csc2 (3)
8
We can apply this result to this problem. For a segment of wire, as set up above, the magnetic eld at a point P is B= 0 I (cos 1 cos 2 ) . 4a
For a square wire loop consider the bottom segment. Using the above general formula for this case gives Bone = 0 I 4 cos 3 cos 4 4
2 0 I = 2, 2 B = 4 Bone 2 0 I 2, =
where
since all four sides contribute and the direction and magnitude of the eld is the same for each side. 012 (part 2 of 2) 10 points What is the direction of the magnetic eld BP at point P due to the downward current in the left-hand side of the square wire? 1. B is zero. 2. B is up the page. 3. B is to the left. 4. B is out of the page. correct 5. B is down the page. 6. B is to the right. 7. B is into the page. Explanation: The direction of the magnetic eld due to a current element is determined by the cross product in the denition of the magnetic eld B= 0 I 4 ds r . 2 r
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban For the present case the right hand rule gives the direction of the magnetic eld as out of the page or out for short. 013 (part 1 of 1) 10 points An unpolarized light beam with intensity of I0 passes through 2 polarizers shown in the picture.
Unpolarized light Polarizer E0 Analyzer E 0 cos
9
=
I0 cos2 (30 ) 2 3 I0 = 8
Transmission axis
Polarized lihgt
If = 30 ,what is the beam intensity after the second polarizer? 5 I0 16 3 2. I0 correct 8 1 3. I0 16 1 4. I0 2 3 5. I0 16 5 6. I0 8 9 7. I0 16 1 8. I0 8 7 9. I0 16 1 10. I0 4 Explanation: The beam intensity after the rst polarizer is I0 I1 = 2 We use the formula for the intensity of the transmitted (polarized) light. Thus the beam intensity after the second polarizer is 1. I = I1 cos2
014 (part 1 of 1) 10 points Four long, parallel conductors carry equal currents I = 6.27 A. An end view of the conductors is shown in the gure. Each side of the square has length = 1.49 m. Note: The current direction is out of the page at points A and B (indicated by the dots) and into the page at points C and D (indicated by the crosses).
A 6.27 A
1.49 m
C +6.27 A
P
6.27 A B
+6.27 A D
Which of the diagrams shown above correctly denotes the directions of the components of the magnetic eld from each conductor at the point P ? Note: BA,B means the magnetic eld from currents A and B, etc. 1. correct 2. BB,C BA,D BB,C BA,D
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban BA 3. P BB 4. BA,D BB 5. L BD BC E C Sb a R BB,C BA BD BC posed on one another. BB,C BA,D
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015 (part 1 of 2) 10 points Consider the following circuit. After leaving the switch at the position a for a long time, move the switch from a to b. There will be the usual LC circuit oscillations.
Explanation: The directions of the magnetic eld due to each wire are given by the right hand rule, where the thumb points in the direction of the current and your ngers curl in the direction of the magnetic elds circular path. Note that the magnetic eld from each wire circulates in a circle around that wire. See the gure for the components from wires A and B. Note: At the point P, you take the magnetic eld direction tangent to the circle formed by magnetic eld lines. A BB 1.49 m BA C 6.27 A
The maximum current will be given by 1. Imax = 2. Imax = E R E R C . L L . C
3. Imax = E 4. Imax
L . C E = LC . R C . correct L 1 . LC LC .
5. Imax = E 6. Imax = E 7. Imax = E
P B D
8. Imax =
E . R
The gure above shows the magnetic eld contributions due to the currents in A and B only. The gure below shows the results of all four magnetic eld contributions superim-
Explanation: Basic Concepts: RCL circuit 12 1 1 2 L Imax = qmax = C E 2 2 2C 2
Husain, Zeena Exam 3 Due: Apr 13 2004, 10:00 pm Inst: Sonia Paban Imax = C E. L
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5T , q is negative, i is counterSo at t = 8 clockwise.
016 (part 2 of 2) 10 points Consider the following statements: A1. The current ow is counterclockwise. A2. The current is zero. A3. The current ow is clockwise. B1. The charge on the left plate of C is positive. B2. The charge on the left plate of C is zero. B3. The charge on the left plate of C is negative. Let the time when the switch is moved from a to b be at t = 0. Which pair of choices below best describes 5 the situation at t = T , T is the period of 8 oscillations in LC circuit. 1. A2 2. A3 3. A2 4. A1 5. A3 6. A1 7. A2 8. A3 9. A1 and B2 and B3 and B3 and B3 correct and B2 and B2 and B1 and B1 and B1
Explanation: At t = 0 left plate of C has q = +qmax , i = 0. T At t = left plate of C has q = 0, i is 4 clockwise. T left plate of C has q = qmax , At t = 2 i = 0. 3T At t = left plate of C has q = 0, i is 4 counterclockwise.