Physics 435 HW 11: Resistor Surface Charge, Mutual Inductance, Solenoid Energy, Assignments of Guiding Electromagnetic Systems

Solutions to homework #11 of physics 435, a university-level course from spring 2008. The homework covers various topics in physics, including finding the resistance and surface charge density of a cylindrical resistor, calculating mutual inductance between two loops, and determining the energy stored in a solenoid's magnetic field. The problems involve applying gauss's law, the neuman formula, and faraday's law.

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Pre 2010

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Physics 435
Spring 2008
Homework #11
1) Consider a resistor built out a cylinder of radius r which carries a current
I
.
The material has a conductivity of
σ
1
over the length 1
followed by a material
with a conductivity 2
σ
over the length 2
. Answer all parts of this problem in
terms of 22
, , , , and rI
σ
σ
11
 . Assume the two electrodes are perfectly
conducting plates of radius r.
a) Find the resistance of the resistor
b) Find the surface charge density between the first electrode and the start of
the
σ
1 material, at the junction of the
σ
1
and 2
σ
material, and between the
2
σ
material and the second electrode. Find the total amount of surface
charge tot
Q in these three regions. Hint-think Gauss’s Law.
c) Is there any volume charge density?
2) Griffiths problem 7.20
3) Consider two loops of radius “a” parallel to the x-y plane. Both centers are on
the z axis but displaced by a distance Z. with
Z
a
a) Calculate the mutual inductance between the two loops by assuming that
one produces a magnetic dipole field at the center of the second loop and
the second loop is so small that the flux is the area of the loop times the
field. This part is a special case of Griffiths 7.20.
b) Now calculate the mutual inductance using the Neuman formula. Here is
how I did it. I parameterized a position vector on either loop by 2
and
φ
φ
1 I
expanded 222
12 12
1/( ) ( )
xyyZ−+−+
to the lowest non-vanishing order in
22
12
()/
x
xZ and 22
12
()/yy Z using a binomial expansion. The mutual
inductance is then a double integral of the form
()
2
2
0
Mdd
ππ
φφ
2
1
0
=∫∫ which
is slightly tedious but do-able.
pf2

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Physics 435

Spring 2008

Homework

  1. Consider a resistor built out a cylinder of radius r which carries a current I.

The material has a conductivity of σ 1 over the length A 1 followed by a material

with a conductivity σ 2 over the length A 2. Answer all parts of this problem in

terms of I , σ 1 , A 1 , σ 2 , A 2 and r. Assume the two electrodes are perfectly

conducting plates of radius r.

a) Find the resistance of the resistor

b) Find the surface charge density between the first electrode and the start of

the σ 1 material, at the junction of the σ 1 and σ 2 material, and between the

σ 2 material and the second electrode. Find the total amount of surface

charge Q tot in these three regions. Hint-think Gauss’s Law.

c) Is there any volume charge density?

  1. Griffiths problem 7.

3) Consider two loops of radius “a” parallel to the x-y plane. Both centers are on

the z axis but displaced by a distance Z. with Z  a

a) Calculate the mutual inductance between the two loops by assuming that

one produces a magnetic dipole field at the center of the second loop and the second loop is so small that the flux is the area of the loop times the field. This part is a special case of Griffiths 7.20.

b) Now calculate the mutual inductance using the Neuman formula. Here is

how I did it. I parameterized a position vector on either loop by φ 1 andφ 2 I

expanded 1 / ( x 1 (^) − x 2 (^) )^2 + ( y 1 (^) − y 2 )^2 + Z^2 to the lowest non-vanishing order in 2 2 ( x 1 (^) − x 2 ) / Z and 2 2 ( y 1 (^) − y 2 ) / Z using a binomial expansion. The mutual

inductance is then a double integral of the form (^) ( )

2 2 0

M d d

π π

2 1 0

= ∫ ∫ which

is slightly tedious but do-able.

Page 2 of 2

4) Consider an long solenoid with radius R and length L that holds a uniform B-

field. You calculate the energy stored in the magnetic field using

2 V S

U B Bd τ A B da μ (^0)

= ⎢ − × ⎥

G G G G G

i i but instead of computing this from the

volume of the solenoid, you decide to use a cylinder of length L and radius “a” coaxial with the solenoid for the volume term and the use surface of the cylinder for the surface. Do you get a reasonable answer? For example compare the answer of a infinitesimally smaller than R to the case where a is infinitesimally larger than R. What went wrong? Hint – In our Potential Lecture Notes, we worked an analogous electrostatic problem using

( ) V S

U E E d V r E da

ε ε = (^0) ∫ τ+^0 ∫ 2 2

G G G G G

i i for a spherical shell of radius R, and charge Q

where we used a spherical volume and surface with a radius d > R. What would have happened if we used d < R instead.

  1. Consider an infinitely long air-core solenoid for radius s = R. The symmetry axis of the solenoid is along the z-axis. A constant current would create a constant B field. In this problem you are asked to compute the electric and magnetic fields when a sinusoidal varying current flows through the solenoid.

The fields can be written as a product of sin ω t or cos ω t and a Taylor series

in increasing powers of

s c

. To zero order the magnetic field is given by 0 0

B^ G^ (^ )^ =ˆ z B sin ω t. You should use the variables , ω

R , Β 0 and physical constants to answer all parts of this problem.

a) Use the zero order B-field expression to obtain a first order (in ω ) expression for Δ E s t ( , )

G

. This is easy to do with Faraday’s law.

b) Using the Δ E ( , ) s t

G

expression you obtained in a) to obtain ( , )

G

B s t to second

order in ω.

c) From the foregoing, the following expansion for the magnetic field ( ) even

( , ) (^) ˆsin

n n n

s B s t z t B c

G

seems plausible. Find a recursion relation

between the coefficients B (^ n ). In this case, the recursion relation shows how to obtain the coefficient B (^ n +^2 )from the lower order coefficient: B (^ n ). The approach is to take the n’th term in the B

G

series, compute Δ E

G

for this term using Faraday’s law and then compute Δ B

G

from the Δ E

G

using the Maxwell displacement current.