Columbia UniversityDepartment of Physics

QUALIFYING EXAMINATION

Monday, January 9, 20173:00PM to 5:00PMClassical Physics

Section 2. Electricity, Magnetism & Electrodynamics

Two hours are permitted for the completion of this section of the examination. Choose 4 problemsout of the 5 included in this section. (You will not earn extra credit by doing an additional problem).Apportion your time carefully.

Use separate answer booklet(s) for each question. Clearly mark on the answer booklet(s) whichquestion you are answering (e.g., Section 2 (Electricity etc.), Question 2, etc.).

Do NOT write your name on your answer booklets. Instead, clearly indicate your Exam LetterCode.

You may refer to the single handwritten note sheet on 812” × 11” paper (double-sided) you have

prepared on Classical Physics. The note sheet cannot leave the exam room once the exam hasbegun. This note sheet must be handed in at the end of today’s exam. Please include your ExamLetter Code on your note sheet. No other extraneous papers or books are permitted.

Simple calculators are permitted. However, the use of calculators for storing and/or recoveringformulae or constants is NOT permitted.

Questions should be directed to the proctor.

Good Luck!

Section 2 Page 1 of 6

1. An infinite straight wire carries zero current at all times except for a sharp pulse at t = 0, i.e.I(t) = I0τδ(t), where δ(t) is the Dirac Delta function and I0τ is a given constant.

(a) Calculate (in Lorentz gauge) the potentials everywhere in space and time

(b) Calculate the electric and magnetic fields everywhere in space and time

Section 2 Page 2 of 6

2. Consider a cylinder of radius R and length L that is uniformly charged with charge density ρ.The cylinder rotates about its axis (the z-axis) with angular velocity ω, as shown in the figure.The top of the cylinder lies in the xy-plane z = 0.

(a) Compute the magnitude of the current density, J, as a function of distance, r, from thecenter of the cylinder.

(b) Compute the magnetic induction, ~B, along the z-axis at the position ~r = (0, 0, h), i.e. atheight h above the top of the cylinder.

(c) Find the expression for Bz(h) in the limit of h � R, L, i.e. very far away from the cylinder.You can solve this directly using your solution to part (b), or by alternative means. Evenan order-of-magnitude estimate, with the correct units, will result in partial credit.

Section 2 Page 3 of 6

3. A point particle of charge q and position ~x(t) moves according to

~x(t) = 0 when t < 0

~x(t) = ~v0

[t −

r0

v0

(1 − e−v0t/r0

)]when t > 0

(1)

where v0 = |~v0|with~v0 a fixed velocity and r0 a fixed length. (The above formulas represent theemission of a charged particle during the decay of a state at t = 0.) Evaluate the total electro-magnetic energy radiated by the particle in the dipole approximation. Under what conditions,on v0 and r0, is the dipole approximation valid?

Section 2 Page 4 of 6

4. Consider a very long coaxial cable. The inner conductor is a cylinder of radius a and carriesfree charge with linear charge density λfree. It is surrounded by a cylindrical linear dielectricwith permittivity ε and radius b.

(a) Find the following quantities for all three regions (i) s < a, (ii) a < s < b, and (iii) b < s:

1. Electric displacement ~D;2. Electric field ~E;3. Polarization ~P.

(b) Calculate the potential V on the axis (s = 0) of the conductor.

(c) Calculate the bound charges, i.e. give the bound volume charge density ρb in the bulkof the dielectric and the bound surface charge density σb on the inner (s = a) and outer(s = b) surfaces of the linear dielectric.

Section 2 Page 5 of 6

5. The figure below shows a spherical hollow inside a uniformly charged sphere of radius R; thesurface of the hollow passes through the center of the sphere and “touches” the right side of thesphere. The charge of the sphere before hollowing was Q. With what electrostatic force doesthe hollowed-out charged sphere attract charge q that lies at a distance d from the center of thecharged sphere, on the straight line connecting the centers of the sphere and of the hollow?

Section 2 Page 6 of 6

Pasupathy Section 2-1

Electricity & Magnetism (Metzger)Consider a cylinder of radius R and length L that is uniformlycharged with charge density ρ. The cylinder rotates with a uni-form angular velocity ω through its center around the z-axis, asshown in the figure. The x-y axis (z = 0) lies parallel to the topof the cylinder.

1. Compute the magnitude of the current density, J, as a func-tion of distance, r, from the center of the cylinder.

2. Compute the magnetic induction, ~B, along the z-axis atthe position ~r = (0, 0, h), i.e. a high h above the top of thecylinder.

3. Simplify the expression for the value of Bz(h) in the limitthat h � R,L, i.e. very far away from the cylinder. Youcan solve this directly using your solution to part 2, or byalternative means. Even an order-of-magnitude estimatehere with the correct units will result in partial credit.

1

Metzger Section 2-2

Solution:

1.~J(r) = ρ~v(r) = ρ(ω × ~r)⇒ | ~J | = ρωr

2.~B =

µ0

4π

∫d3~r′ ~J(~r′)× ~r − ~r′

|~r − ~r′|3

and hence, defining an angle α between the source point ~r′ = (r′, φ′) and the locationon the axis ~r = zz,

Bz(~r) =µ0

4π

∫d3~r′J(~r′)

sinα

|~r − ~r′|2=µ0

4π

∫ L

0

dz′∫ 2π

0

dφ′∫ R

0

r′dr′ρωr′r′

[(h+ z′)2 + (r′)2]3/2

=µ0

2ρω

∫ L

0

dz′∫ R

0

dr′(r′)3

[(h+ z′)2 + (r′)2]3/2=µ0

2ρω

∫ L

0

dz′

[2(h+ z′)2 +R2

[(h+ z′)2 +R2]1/2− 2(h+ z′)

]

=µ0

2ρω[(h+ L)

(√(h+ L)2 +R2 − (h+ L)

)− h

(√h2 +R2 − h

)]3. There is an easy and a hard way to solve this problem.

First, we can expand Bz in terms of small parameters h/L and h/R,

Bz =µ0

2ρωh2

[(1 + L/h)

(√1 + 2L/h+ (L/h)2 + (R/h)2 − (1 + L/h)

)−(√

1 + (R/h)2 − 1)]

Using the expansion (1 + ε)1/2 ' 1 + ε/2− ε2/8 + ε3/16− 5128ε4 for ε = R/h, L/h� 1,

after much simplification, we obtain

Bz 'µ0

2ρωh2 × LR5

4h5=µ0LR

4

8h3ρω

As a second, easier approach, we note that far away from a magnetic dipole ~m = mz,the B-field along the direction z is

Bz 'µ0

2π

m

z3=z=h

µ0LR4

8h3ρω,

where

|~m| = 1

2

∫|~r′ × ~J |d3~r =

1

2

∫ 2π

0

dφ

∫ R

0

r · (ρrω) · rdr = πLρω

∫ R

0

r3dr =πLR4

4ρω

is the magnetic moment.

2

Metzger Section 2-2

MuellerSection 2-3

Karagiorgi Section 2-4

Karagiorgi Section 2-4

Karagiorgi Section 2-4

Johnson Section 2-5