Answers to Old Electromagnetics Exams

Please let me know if you have problems with any of these, and feel free to ask for details. --Ruth Douglas Miller, 11 April 2006.

Exam I (Vector calculus, electrostatics)

a) and b) 8.99e7/r^2 V/m in r direction. c) Q = -10mC evenly distributed on inside of shell; Q = 10mC evenly distributed on outside of shell (you figure out why!)
T = r (theta direction) + r sin(theta) cos(phi) (phi direction)
1.8e5/rho V/m
4.14e5 V!
-2xy (x) - (x^2 + 2yz) (y) - y^2 (z) V/m (x) indicates unit vector
4 uC
B and C are not valid electrostatic fields: curl E is not zero. (Note typo on C: y component should be xz^2 - 2 yz
ans should be 2, both ways.

F = 6.4e-13 N
We=2.25e-18 J
Can assume tree, being a conductor, though poor, is all at one (ground) potential. The shortest path to ground for a charge in the atmosphere is thru the tallest tree, because the attractive force btwn the charge and ground is greatest where the potential change is greatest.
Flux lines form arcs from sides of needle to ground; equipotentials form approximate hyperbolas with centerlines coincident with needle centerline.
D = 15e-9/2r in the r direction, where r = rho.
E = 13.5 kV/m along x.
I = dQ/dt. A slow-moving Q entering the depletion region feels a strong force (see #6) and is accelerated; thus a small I entering is transformed to a large one (fast-moving Q) leaving the depletion region.

-7190 x + 3.9e9 z (V/m)
whirlpools, eddies: circular water movement
a) 0. b) 0 c) 1.6e-19 - 8e-10*R d) ans. for (c) divided by 4*pi*R^2
circulation of F = line integral of F about a closed path (F is a vector field)
KVL: sum of voltage rises/drops around a closed path = 0. Voltage change = line integral of E from A to B, so KVL suggests line integral of E around a closed path = 0; that is, E is curl-free, irrotational, conservative.
a) sqrt(29) b) 2x - 5y or parallel to that vector
(BONUS): fact: an observation detectable by our 5 senses (though potentially thru instruments we read with our 5 senses.) Theory: an explanation of a (large) set of facts/observations that also predicts the outcome of future tests/observations, and has been found extremely reliable with many many observations. Law: another word for theory; may imply that the theory may be stated in mathematical terms (an equation).

66.6 degrees
8.35e-14 F (1/2 a sphere)
a) 70.8 ns = 5 time constants. b) same time
1.2
inside inner conductor H = 795r A/m. Btwn conductors H = 7.96e-4/r A/m. Outside both conductors H = 0.
Add a second, larger thin coil at each end of the solenoid, carrying a current in the reverse direction. (other answers possible)
E = 21.8 kV/m

Rinner = 0.125 ohms, Router = 0.0269 ohms
C = 1 uF by time constant method, 1.56 nF by formula
5 time constants = 50 sec; charge on inner moves to outer and neutralizes
b: loops of current around bar. c: loop current I within Earth, in equatorial plane or parallel to it
electrons are torn free of their atoms (ionize) and are then free to conduct--material becomes conductor
E = 0, r <a. E = rho-naught*a^2* (r-a)/(epsilon*r^2)
more dipoles/vol (denser); larger dipoles/molecules, stretchier molecules all increase permittivity. Note conductivity and permittivity and dielectric strength are all independent of eachother!
since number and charge of carriers hasn't changed, mobility must have increased dramatically.

1.8e-12 m
5e-16 S/m
zero: Al is conductor and shields interior from E fields.
a) 10^-6 C b) 2e-6 C/m^2 c) 9e3/r^2 V/m d) rho(s) R/epsilon = 9e4 V e) 4*pi*epsilon *R = 1.11e-11 F.
3.36e9 V/m (independent of Q)
consider the direction of force on positive or negative charge carriers moving at right angles to an applied B field. THe polarity of the material perpendicular to the applied voltage will be opposite for e-s than it is for holes.
BONUS: it won't; the long-wavelength AM waves will not penetrate the house any more easily than 60Hz does.

flux = 1.1e-4 Wb; m = 3.3 kg
M = (N1 N2 A u1 u2)/2*pi*R (avg core radius R, cross-sectional area of core A)
Vemf causes current to flow to create a B that opposes the changing applied B. In a s/c that current perfectly cancels the applied B as it increases as a magnet is brought near, b/c the current faces no resistance in the s/c.
F = 0
Vemf= 0.94V
Es=6 e^(0.3x) e^(j5x) V/m in z. Polarization linear in z, atten const 0.3, phase const 5, wavelength 1.26, freq 1 MHz, velocity 1.26e6 m/s, travels in -x direction.
150 MHz
(note first integral should be negative and closed surface; 3rd & 4th integrals need d/dt in front): total power entering closed surface S = power dissipated in conductive medium (resistive loss) + increase in time of energy stored in elec dipoles in volume (capacitive energy) + increase w/time of energy stored in mag dipoles in volume (inductive energy)
walls and soil are good conductors at frequency of cell phone operation; signal penetrates only to skin depth and is absorbed.

inner conductor: H = 1591r A/m. Btwn conductors, H = 1.59/r A/m. Outside both conductors H = 0.
current flows CCW as given. If loop moves away at the right speed, I2 will reverse direction as flux decreases.
flux = 2.62e-4 Wb; m = 5.57 kg
the axle of the disk is positive relative to the rim.
V = w B (b^2 - a^2)/2 where w is omega.
a pair of concentric coils spaced a distance apart; a solenoid, a spherical solenoid...

a) 0.4pi(a^2/2 + a + 1 -e^(-a)) A. b) H = 0.2(r/2 + e^(-a)/r (e^r (r+1) -1)) A/m
I2= 5.27e-10 A, CW
force on wires (and cone) is I x B force; speaker moves up and down as I follows sound freq and intensity
magnitude 700 V/m, direction -y, freq 800 MHz, phase const 83.8, atten const 0, phasor 700 e^(j83.8y) along z.
180 kHz
given displacement term, it is clear that E creates B and B creates E: no medium is needed to contain energy.
10 uV
(BONUS): determine velocity of coming train before it is within sight. Check speedometer of train. (assumes R not too high.)