PHYS C1402(2) Midterm 2 Exam
Thursday, April 2"d, 2009
Prof. Jeremy Dodd
Answer all three questions. Point allocations for each question are shown in parentheses. You may
answer the questions in any order, but start each problem on a new page. Be sure to
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Introduction to Electricity, Magnetism, and Optics
C1402 (Section 1)
Spring 2007
Final Exam (05/07/07)
There are 170 minutes permitted for the complete examination. Use the exam
book for your answers. Do not discuss the exam at any time. The exam is
close
Introduction to Electricity, Magnetism, and Optics C1402
Spring 2006
Midterm 2 (03/28/06)
There are 75 minutes permitted for the complete examination. Use the exam
book for your answers. Do not discuss the exam at any time. The exam is
closed book, but yo
PHYS C1402(2) Final Exam
Tuesday, May 12th, 2009
Prof. Jeremy Dodd
Answer all eight questions. Point allocations for each question are shown in parentheses. You may
answer the questions in any order, but start each problem on a new page. Be sure to show a
PHYS C1402(2) Midterm 1 Exam
Thursday, February 19th, 2009
Prof. Jeremy Dodd
Answer all three questions. Point allocations for each question are shown in parentheses. You may
answer the questions in any order, but start each problem on a new page. Be sure
Assignment 1 Solutions
Chpt 21 8. For ease of presentation (of the computations below) we assume Q > 0 and q < 0 (although the final result does not depend on this particular choice). (a) The x-component of the force experienced by q1 = Q is - ( Q
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4. Problem 23.42. The surface charge density is given by
E / 0 0 E 8.85 1012 C2 /N m2 (55 N/C) 4.9 1010 C/m2 .
Since the area of the plates is A 1.0 m2 , the magnitude of the charge on the plate is
Q A 4.9 1010 C.
5. Problem 23.54. Applying Eq. 23-20, we
Introduction to Electricity/Magnetism/Optics, 1402, Spring 2015
Final Exam.
Tuesday, May 12, 9:00 am.
You should complete all eight problems (25 points each). Some useful information is included
on the last five pages. Good luck!
1. Insect inside an amber
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4. Problem 21.50. (a) Since the rod is in equilibrium, the net force acting on it is zero, and the net
torque about any point is also zero. We write an expression for the net torque about the bearing,
equate it to zero, and solve for x. The charge Q on th
4. Problem 24.10. In the inside region between the plates, the individual fields (given by Eq. 24-13)
are in the same direction ( i ):
50 109 C/m2
25 109 C/m2
Ein
i (4.2 103 N/C)i .
12
2
2
12
2
2
2(8.85 10 C /N m ) 2(8.85 10 C /N m )
In the outside re
Do all problems. Show all your work if you want partial credit. There is an equation sheet attached.
Good‘luck and have a nice summer!
Problem 1 (40)
Assume a point-like nucleus of charge Q: +Ze where Z 2 atomic number is surrounded by a cloud of
electron
Chapter 28: Magnetic Fields
28.1 Magnetic Fields and the Definition of
o What Produces a Magnetic Field
Electromagnet, the current produced can be used to control a computer hard drive,
etc.
Permanent magnet, has permanent magnetic field
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Physics C1402 Formula Sheet
i
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p
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Charges Q) +27, and i are Posi-Eomd as shown.
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4. Problem 28.72. (a) For the magnetic field to have an effect on the moving electrons, we need a non
negligible component of B to be perpendicular to v (the electron velocity). It is most efficient, therefore,
to orient the magnetic field so it is perpen
4. Problem 26.61. The amount of charge that strikes the surface in time t is given by q = i t,
where i is the current.
Since each alpha particle carries charge q = +2e, the number of particles that strike the surface is
N
q it
.
2e 2e
For part (b), let N
4. Problem 25.50. Let
C1 = 0(A/2)1/2d = 0A1/4d,
C2 = 0(A/2)2/d = 0A2/2d,
C3 = 0A3/2d.
Note that C2 and C3 are effectively connected in series, while C1 is effectively connected in parallel with
the C2-C3 combination. Thus,
C C1
C2C3
A A d 2 2 3 2 0 A
2
4. Problem 29.41. The magnitudes of the forces on the sides of the rectangle that are parallel to the
long straight wire (with i1 = 30.0 A) are computed using Eq. 29-13, but the force on each of the sides
lying perpendicular to it (along our y axis, with
4. Problem 31.48. (a) With both switches closed (which effectively removes the resistor from the
circuit), the impedance is just equal to the (net) reactance and is equal to
Xnet = (12 V)/(0.447 A) = 26.85 .
With switch 1 closed but switch 2 open, we have
4. Problem 30.57. (a) Before the fuse blows, the current through the resistor remains zero. We apply
the loop theorem to the battery-fuse-inductor loop: L di/dt = 0. So i = t/L. As the fuse blows at t = t0,
i = i0 = 3.0 A. Thus,
t0
i0 L
3.0 A 5.0 H 1.5
4. Problem 26.48. The mass of the water over the length is
m AL (1000 kg/m3 )(15 105 m2 )(0.12 m) 0.018 kg ,
and the energy required to vaporize the water is
Q Lm (2256 kJ / kg)(0.018 kg) 4.06 104 J .
The thermal energy is supplied by joule heating of the
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