Description
1) Multipole expansion:
Consider the second charge distribution considered in the Multipole Expansion workshop:
q at (x = 0, y = 0, z = a), q at (x = 0, y = 0, z = ‐a), ‐2q at (x = 0, y = 0, z = 0)
a) Using the multipole expansions for the potential 𝜙ሺ𝑟⃗ሻ, calculate the third (quadrupole) term
in for 𝜙ሺ𝑟⃗ሻ for this charge distribution.
b) Write your result in spherical coordinates, 𝑟, 𝜃,𝜙. Explain why your result doesn’t depend on
𝜙. In this, quadrupole, approximation, for what directions in space is the potential equal to
zero? (Draw a picture.)
c) Using spherical coordinates, calculate the electric field of this charge distribution in the
quadrupole approximation. Sketch the field in the x‐z plane.
2) Charge‐Dipole interaction:
Consider a point charge 𝑄 at the origin 𝑟⃗ ൌ 0 and an electric point dipole 𝑝⃗ at a position 𝑟⃗
(not at the
origin).
Calculate:
a) The potential energy of the 𝑄 & 𝑝⃗ system, by:
i. Calculate the potential energy of 𝑝⃗ in the electric potential of 𝑄
ii. Calculate the potential energy of 𝑄 in the electric potential of 𝑝⃗.
iii. Show these are equal.
b) The forces on 𝑝⃗ and 𝑄, by:
i. Calculate the force on 𝑝⃗ due to the field of 𝑄
ii. Calculate the force on 𝑄 due to the field of 𝑝⃗
iii. Compare the results.
3) Magnetic Force on a Loop:
Consider the two currents shown: (i) a circular loop of radius 𝑅 with
counterclockwise current 𝐼ଵ centered at the origin in the x‐y plane
and (ii) a very long straight wire with current 𝐼ଶ parallel to the y‐direction at
𝑧 ൌ 0, 𝑥 ൌ െ𝑑.
We’re going to find solutions for the force on the loop, first by integration
and then by considering the potential energy.
a) Predict the direction of the total force on the loop due to the
magnetic field of the long‐straight wire. Explain your prediction.
b) What is the magnetic field 𝐵ሬ⃗
ଶሺ𝑟⃗ሻ (magnitude and direction) due
to the long wire everywhere in the x‐y plane?
𝑥
𝑦
𝜙
𝐼ଵ, 𝑅
𝐼ଶ
c) The force on the loop due to the magnetic field of the wire is:
𝐹⃗
ଶଵ ൌ 𝐼ଵ ර 𝑑𝑙⃗ ൈ 𝐵ሬ⃗
ଶ
It should be clear that we want to do this integral over the circle by integrating over 𝑑𝜙.
i) Derive expressions for 𝐵ሬ⃗
ଶሺ𝑟⃗ሻ and 𝑑𝑙⃗in terms of 𝑅,𝜙, 𝑑, and 𝑑𝜙 (for points on the
current loop). Write these in terms of the 𝑥ො and 𝑦ො components.
Check your answer at a few simple points (such as for the angles 𝜙 ൌ 0,𝜙 ൌ గ
ଶ , … ሻ
iii) Write out an integral that gives the force on the loop. You should simplify this as
much as possible, but you don’t need to solve it, as it’s somewhat messy.
Does the direction agree with your prediction?
d) Another approach to this problem is to determine the potential energy of the loop due to the
magnetic field of the wire, 𝑈, and then use that the force is 𝐹 ൌ െ∇ሬሬ⃗ 𝑈. In this case
you’re interested in the change in energy as the loop moves relative to the wire
𝐹௫ ൌ െ𝜕ௗ 𝑈
i) A current loop is equivalent to a sheet of small current loops, each a magnetic dipole
(Fig. 4.7 in the textbook). Using small loops, 𝑑𝑆⃗ ൌ 𝑑𝐴 𝑛ො, gives magnetic dipoles
𝑑𝑚ሬሬ⃗ ൌ 𝐼 𝑑𝐴 𝑛ො
The potential energy of the magnetic dipoles
𝑑𝑈ሺ𝑟⃗ሻ ൌ െ𝑑𝑚ሬሬ⃗ ⋅ 𝐵ሬ⃗
ଶሺ𝑟⃗ሻ
Write a surface integral for the potential energy of the loop. Use polar coordinates.
ii) Write the field 𝐵ሬ⃗
ଶሺ𝑟⃗ሻ everywhere inside the loop using polar coordinates. (This is a
simple extension to part c above.)
iii) Solve your integral to determine 𝑈 and take the derivative to get 𝐹௬. Does your
result make sense? It might be useful to know that:
න 𝑑𝜙
𝑑𝑟 cos 𝜙
ଶగ
ൌ 2𝜋
√𝑑ଶ െ 𝑟ଶ
iv) Consider your result for the force on the loop in the limit that 𝑑≫𝑅. Show that this
is equivalent to the force on a magnetic point dipole due to the magnetic field of the
wire. Remember that the force on a dipole is
𝐹⃗ ൌ ∇ ሺ 𝑚ሬሬ⃗ ⋅ 𝐵ሬ⃗ ሻ
4) Back to the Ring:
Consider one more time the current‐carrying ring considered in class.
The ring is in the x‐y plane with radius 𝑎 and a counter‐clockwise
current 𝐼. (Bottom picture)
As before, we want to calculate the magnetic field at a point
𝑟⃗ൌ𝑟 cosሺ𝜃ሻ 𝑧̂ 𝑟 sinሺ𝜃ሻ 𝑥ො
In this case we’ll look at this problem using a multipole expansion in
spherical harmonics. We’ll use the result:
1
|𝑟⃗െ𝑟⃗ᇱ|
ൌ 4𝜋
2𝑙 1
𝑎
𝑟ାଵ 𝑌
∗ ሺ𝜃ᇱ
,𝜙ᇱ
ሻ 𝑌ሺ𝜃,𝜙ሻ
ୀ ି
ஶ
ୀ
a) Write down (or look up from class) and expression for the
current density of the ring, 𝐽⃗ሺ𝑟⃗ᇱ
ሻ in terms of the spherical
coordinates 𝑟ᇱ
, 𝜃ᇱ
,𝜙′.
Remember that for a 1D distribution you’ll need two 𝛿‐functions.
b) Using the definition for the vector potential:
𝐴⃗ሺ𝑟⃗ሻ ൌ 𝜇
4𝜋 ම𝐽⃗ሺ𝑟⃗ᇱ
ሻ 1
|𝑟⃗െ𝑟⃗ᇱ| 𝑑ଷ𝑟′
Write down the multipole (spherical harmonics) expansion for the vector potential. This should
still include sums over 𝑙 and 𝑚 and the volume integral over 𝑟⃗′. Remember to include the vector
direction(s) of the current density.
c) Perform the integrals over 𝑑𝑟′ and 𝑑𝜃′ to determine an expression for 𝐴⃗ሺ𝑟⃗ሻ just in terms of an
integral on 𝑑𝜙′. Of course, you’ll still have the sums over 𝑙 and 𝑚.
d) Show that the 𝑙 ൌ 0 term in the expansion is zero.
e) Calculate the 𝑙 ൌ 1 term in the multipole expansion. This will include the sum over
𝑚 ൌ 0, േ1.
f) Calculate the magnetic field 𝐵ሬ⃗ ൌ ∇ሬሬ⃗ ൈ 𝐴⃗. Compare your results to those found in class,
including the results for 𝐵ሬ⃗ሺ𝑟⃗ൌ𝑧 𝑧̂ሻ and for 𝑟≫𝑎.
For Fun) Try calculating the 𝑙 ൌ 2 term in the multipole expansion.
𝑥
𝑦 𝑰, 𝒂
𝒓ሬ⃗′
𝝓′
𝑥
𝑧 𝒓ሬ⃗
𝒓ሬ⃗ െ 𝒓ሬ⃗′
𝒓ሬ⃗′
𝑦
𝜽
5) Magnetic Dipole Interactions in a Magnetic Field:
NOTE: In this problem, you can assume for all your answers that
𝒎ሬሬሬ⃗𝟏and 𝒎ሬሬሬ⃗𝟐 are either parallel or anti‐parallel. I’m fairly sure that this
has to be true, but there are still a couple of cases where I need a
more complete proof. If you want to prove this, please do. If not, you
can assume it.
Two identical magnetic dipoles are shown, 𝑚ሬሬ⃗ଵ at the origin and 𝑚ሬሬ⃗ଶ at
𝑟⃗ൌ𝑑 𝑧̂. The magnetic dipoles are free to rotate in the x‐z plane (they don’t have components in 𝑦ො).
To
simplify the solution to this problem, define a quantity related to the magnetic field due to the dipoles:
𝐵ௗ ൌ 𝜇
4𝜋
𝑚
𝑑ଷ , |𝑚ሬሬ⃗ଵ| ൌ |𝑚ሬሬ⃗ଶ| ൌ 𝑚
There is a uniform, constant magnetic field 𝐵ሬ⃗ in the x‐z plane.
a) Write down an expression for the total potential energy of the two dipoles interacting with
each other and the magnetic field.
b) First, consider the case where 𝐵ሬ⃗ ൌ 0. What are the are the configurations of the dipoles that
give the lowest potential energy (there are two)? Show that adding a magnetic field 𝐵ሬ⃗ ൌ 𝐵௫௧ 𝑧̂
will result in one configuration having the lowest energy.
c) If instead the magnetic field is 𝐵ሬ⃗ ൌ 𝐵௫௧ 𝑥ො, the lowest energy configuration of the dipoles will
depend on the magnitude of the field.
i) If 𝐵௫௧ ≪ 𝐵ௗ what do you expect the lowest‐energy configuration of the dipoles to
be? If 𝐵௫௧ ≫ 𝐵ௗ what do you expect the lowest‐energy configuration of the dipoles to
be? Explain.
ii) Assuming 𝑚ሬሬ⃗ଵ and 𝑚ሬሬ⃗ଶ remain parallel, determine the lowest energy configuration of
the two dipoles as a function of the magnitude 𝐵௫௧.
Show that there is a “critical” value, 𝐵, for the external field where the lowest‐energy
configuration changes abruptly to the dipoles being aligned with 𝐵ሬ⃗.
𝑧
𝑥
𝑚ሬሬ⃗ଵ
𝑚ሬሬ⃗ଶ
𝑑
