Phys 742 - EM I Final exam - 10 May 1995

1.

(a) Find the self-inductance of a solenoid of length l, consisting of N turns of wire around a core having cross-sectional area A and permeability μ. Neglect the fringing field.

(b) Can you estimate the contribution of the fringing field to the self-inductance L of a long solenoid and come up with a rough simple formula for it? It is convenient to put L = L_{0}f, where L_{0} is the result of part (a) and f is a correction factor. An elaborate argument will earn extra points (leave it for last).

2.

What are the non-electromagnetic dimensions (or units) of the following quantities:

What are the non-electromagnetic dimensions (or units) of the following quantities:

(a) LC, where L is inductance and C capacitance

(b) RC, where R is resistance

(c) RI^{2}, where I is current

(d) ϵE^{2}, where ϵ is dielectric constant, E electric field

(e) μH^{2}, where μ is magnetic permeability, H magnetic field.

3.

(a) What happens, qualitatively, when a charged body is brought near an insulated, uncharged body? Is there repulsion or attraction? How does the force vary with distance at large separations?

(b) Find the force between a point charge q and an uncharged, insulated sphere of radius a. For full credit you are supposed to solve the problem for a sphere of dielectric constant ϵ. If you cannot do that, obtain the answer for a conducting sphere. If you can do the general case, discuss the conducting sphere as a particular case.

4.

A plane electromagnetic wave of angular frequency ω is incident perpendicularly on the surface of a non-magnetic medium of dielectric constant ϵ. To keep the notation uniform, assume that the wave propagates in the z direction and that the incident electric field is in the x direction (or use vector methods that are coordinate-independent).

(a) Find the reflected electric and magnetic fields.

(b) Find the electric and magnetic fields in the medium.

(c) Find the reflection coefficient R, i. e. the ratio of reflected to incident intensity.

(d) Find the transmission coefficient T, i. e. the ratio of transmitted to incident intensity, by computing the Poynting vector just inside the medium.

(e) Do you expect that R + T = 1 for real ϵ? For complex ϵ? Check that the results of (c) and (d) agree with your expectations.

5.

Same as problem 4, but for a conductor characterized by a real conductivity σ. The easiest way is to note that a real σ corresponds to a complex ϵ(ω). If you can do that, you do not need to repeat the answers (a) through (e). In addition,

(f) Compute the current density j near the surface and the integrated surface current K.

(g) Compute the dissipated power in the conductor and show that it is consistent with the values of R and T that you have obtained by computing the Poynting vector.

6.

The spectral function of a material is approximately

ω Im ϵ(ω) = \{ \begin{matrix}S & for ω_{1} < ω < ω_{2} (with ω_{1} > 0) \\ 0 & for all other positive ω \end{matrix}

(a) Determine ϵ(ω) for all real values of ω, both positive and negative.

(b) How is S related to the electron density in the material?

\frac{1}{|x − x′|} = ∑_{l} \frac{r_{<}^{l}}{r_{>}^{l+1}} P_{l}(cos θ)

Re ϵ(ω) = 1 + \frac{2}{π} ℙ ∫_{0}^{∞} \frac{ω′ Im ϵ(ω′)}{ω^{′ 2} − ω^{2}} dω′

SOLUTIONS

1.

(a) From Ampère’s circuitation law, lH = 4πNI/c. The energy is

\frac{1}{2} L_{0}I^{2} = \frac{μH^{2}}{8π} Al = \frac{2πN^{2}I^{2}μA}{lc^{2}}

L_{0} can be obtained from this. Equivalently, one can use L_{0}I = (1/c)NμHA, where NμHA is the “linked flux” of B = μH through N turns. Either way:

L_{0} = \frac{4π}{c^{2}} \frac{N^{2}μA}{l}

For the MKS result, omit the factor 4π/c^{2}.

(b) End effects decrease the inductance, as can be seen from the following argument. If the solenoid is broken in two, a new North pole appears on one piece, facing a new South pole on the other. (Positive) work must be done to separate the two pieces by a large distance. This means that the inductance of the unbroken solenoid is greater than the sum of the inductances of the two pieces (remember that \frac{1}{2} LI^{2} tends to a maximum at constant I ).

The reduction of L comes from the weakening of the field inside the solenoid, near the ends. One expects that the correct inductance is L_{0}f, where f − 1 ≈ − a/l, where a is a characteristic size of the solenoid’s cross section. For a round solenoid, a will be the radius (or the diameter), for a square solenoid the side. A simple estimate can be given for a round solenoid when μ = 1. In this case the field on axis is easily calculated (Jackson, problem 5.2) and f can be estimated from the reduction of H^{2}, assuming that the field off-axis scales in the same way. This simple argument gives f = 1 − a/l, while the correct result is f = 1 − (8/3π)a/l, both to leading order in a/l. See Appendix for details.

2.

LC is time squared. RC is time. RI^{2} is power. The last two are energies per unit volume.

(a) There is attraction. At large r, the charge induces a dipole ∼ 1/r^{2}. The force on this dipole is ∼ 1/r^{3}. Thus the force on the body is ∼ 1/r^{5}

(b) The force on a dielectric sphere of radius a is

F (r) = −q^{2} ∑_{l=1}^{∞} \frac{a^{2l+1}}{r^{2l+3}} \frac{l(l + 1)(ϵ − 1)}{ϵl + l + 1}

For a conductor this reduces to

F (r) = −q^{2} ∑_{l=1}^{∞} \frac{a^{2l+1}}{r^{2l+3}} (l + 1) = −q^{2} \frac{a^{3}}{r^{3}} \frac{2r^{2} − a^{2}}{(r^{2} − a^{2})^{2}}

The last expression can be obtained directly by the method of images. For large r/a, in agreement with part (a),

F = −q^{2} \frac{2(ϵ − 1)}{ϵ + 2} \frac{a^{3}}{r^{5}}

4.

(a) With k = ω/c, k′ = n′ω/c, and n′ = \sqrt{ϵ}, the incident fields are:

E_{x} = E_{0} exp (ikz − iωt)
H_{y} = E_{0} exp (ikz − iωt) ,

the reflected fields are

E′′_{x} = − \frac{n′ − 1}{n′ + 1} E_{0} exp (−ikz − iωt)
H′′_{y} = \frac{n′ − 1}{n′ + 1} E_{0} exp (−ikz − iωt)

and the transmitted fields are

E′_{x} = \frac{2}{n′ + 1} E_{0} exp (ik′z − iωt)
H′_{y} = \frac{2n′}{n′ + 1} E_{0} exp (ik′z − iωt)

They satisfy the continuity equations E_{x} + E′′_{x} = E′_{x} and H_{y} + H′′_{y} = H′_{y}.

The (time-averaged) incident intensity is

I_{0} = \frac{c |E_{0}|^{2}}{8π}

The reflected intensity is RI_{0} and the transmitted intensity is T I_{0}, where R and T are the reflection and transmission coefficients:

R = |\frac{n′ − 1}{n′ + 1}|^{2}
T = \frac{4 Re n′}{|n′ + 1|^{2}}

It is always true that R + T = 1, even for complex ϵ.

5.

In a conductor ϵ(ω) = 1 + 4πiσ/ω. The induced current density is

j_{x} = σE′_{x} = \frac{2σ}{n′ + 1} E_{0} exp (ik′z − iωt)

The total surface current is

K_{x} = ∫_{0}^{∞} j_{x} dz = \frac{2σ}{n′ + 1} \frac{E_{0}e^{−iωt}}{α/2}

where α is the attenuation constant defined as in Jackson’s (7.53)

\frac{α}{2} = \frac{ω}{c} Im n′

The dissipated power per unit area of surface is

P = \frac{1}{2} ∫_{0}^{∞} σ |E′_{x}|^{2} dz = \frac{1}{2} \frac{4σ |E_{0}|^{2}}{|n′ + 1|^{2}} \frac{1}{α}  (0.1)

From n^{′ 2} = ϵ, the real and imaginary parts of n′ satisfy the equations

(Re n′)^{2} − (Im n′)^{2} = 1  (0.2)
2 Ren′ Im n′ = \frac{4πσ}{ω}

The last equation can also be written

Ren′ = \frac{4π}{c} \frac{σ}{α}

Using this in (0.1), we see that P = T I_{0} as expected.

The explicit solution of the system (0.2) is

Re n′ = \frac{1}{\sqrt{2}}\sqrt{\sqrt{1 + (\frac{4πσ}{ω} )^{2}} + 1}

Im n′ = \frac{1}{\sqrt{2}}\sqrt{\sqrt{1 + (\frac{4πσ}{ω} )^{2}} − 1}

In particular, for a good conductor at low frequencies (for a metal when ω ≤ 10^{12} sec^{−1}):

Re n′ = Im n′ ≃ \sqrt{\frac{2πσ}{ω}}

Instead of α one could use the skin depth δ = 2/α. For a good conductor

\frac{α}{2} ≃ \frac{1}{c}\sqrt{2πσω} δ = \frac{c}{\sqrt{2πσω}}

6.

(a) Im ϵ(ω) is equal to S/ω for ω_{1} < ω < ω_{2} and for −ω_{1} < ω < ω_{2} and is zero elsewhere. From Kramers-Kronig:

Re ϵ(ω) = 1 + \frac{S}{πω} (ln |\frac{ω_{2} − ω}{ω_{1} − ω} | − ln |\frac{ω_{2} + ω}{ω_{1} + ω} |)

Here is a plot of Re ϵ(ω) versus ω when we take S = π, ω_{1} = 1 and ω_{2} = 2.

(b) Directly from the sum rule:

\frac{4πNe^{2}}{m} = \frac{2}{π} ∫_{0}^{∞} ω Im ϵ(ω) dω = \frac{2S}{π} (ω_{2} − ω_{1})

Equivalently, one can check that for ω → ∞

ϵ(ω) → 1 − \frac{2S}{πω^{2}} (ω_{2} − ω_{1})

and compare this with the general formula ϵ(ω) → 1−ω_{p}^{2}/ω^{2}, with ω_{p}^{2} = 4πNe^{2}/m.

Appendix A

Contents:

– (A) Field on the axis of a finite circular solenoid (Jackson’s problem 5.2) and rough estimate of end-effects for μ = 1 (Smythe)

– (B) Exact self-inductance of a finite circular solenoid for μ = 1 (Grover, page 142).

– (C) Exact self-inductance of a finite rectangular solenoid with μ = 1 (Grover, page 70).

(A) Field on the axis of a circular solenoid

Directly from Ampère’s law, the field on the axis of a circular loop of radius a located in the plane z = z′ is

B_{z}^{loop}(z, z′) = \frac{I}{c} \frac{2πa^{2}}{[a^{2} + (z − z′)^{2}]^{3/2}}

For the solenoid, this must be multiplied by N/l, which is the density of loops, and integrated from 0 to l. Using

∫ \frac{a^{2}dz′}{[a^{2} + (z − z′)^{2}]^{3/2}} = \frac{z′ − z}{\sqrt{a^{2} + (z − z′)^{2}}}

the result is

B_{z}(z) = \frac{2πNI}{lc} (\frac{l − z}{\sqrt{a^{2} + (l − z)^{2}}} + \frac{z}{\sqrt{a^{2} + z^{2}}} )

This agrees with Jackson’s problem 5.2, apart from notation (Jackson’s N corresponds to our N/l). Note that near the middle of a long solenoid (z ∼ l/2) we recover the elementary result B_{z} = 4πNI/lc and that on the end faces B_{z} is half this value.

To find approximately the self-inductance of the solenoid, we use the “linked flux” argument. The flux through one loop is approximately πa^{2}B_{z}(z). Multiplying by N/l and integrating over z from 0 to l we obtain the total linked flux Φ. Using LI = Φ/c we find:

L ≃ \frac{2πN}{lc^{2}} \frac{2Nπa^{2}}{l} (\sqrt{l^{2} + a^{2}} − a)

This result is found in Smythe, page 340. Putting L = L_{0}f we find that the correction factor is

f ≃ \sqrt{1 + (a/l)^{2}} − a/l ≃ 1 − a/l + \frac{1}{2}(a/l)^{2}

The exact result to this order is f = 1 − (8/3π)(a/l) + \frac{1}{2}(a/l)^{2}. See Eq. (0.11) below.

(B) Self-inductance of a finite circular solenoid

It is convenient to use the formula

LI^{2} = \frac{1}{c} ∫ J · A d^{3}x = \frac{2πa}{c} \frac{NI}{l} ∫_{0}^{l} A_{ϕ}(z) dz

where

A_{ϕ}(z) = \frac{NI}{lc} ∫_{0}^{l} dz′ ∫_{0}^{2π} \frac{a cos ϕ′ dϕ′}{\sqrt{2a^{2}(1 − cos ϕ′) + (z − z′)^{2}}}

This vector potential is obtained by integrating over z′ the expression for a circular loop, Jackson’s eq. (5.36) with r sin θ = a and r cos θ = z − z′.

We must compute a triple integral. Putting for short

l^{2}p^{2} = 2a^{2}(1 − cos ϕ′)  (0.3)

it is best to compute first

D(p) = \frac{1}{l} ∫_{0}^{l} dz ∫_{0}^{l} \frac{dz′}{\sqrt{l^{2}p^{2} + (z − z′)^{2}}} = ∫_{0}^{1} dz ∫_{0}^{1} \frac{dz′}{\sqrt{p^{2} + (z − z′)^{2}}}

and then

L = \frac{4πN^{2}a^{2}}{lc^{2}} ∫_{0}^{2π} D cos ϕ′ dϕ′

Integrating by parts over z′:

D = ∫_{0}^{1} dz [\frac{1}{\sqrt{p^{2} + (z − 1)^{2}}} − ∫_{0}^{1} z′ (\frac{∂}{∂z′} \frac{1}{\sqrt{p^{2} + (z − z′)^{2}}} ) dz′]

Replacing ∂/∂z′ with −∂/∂z and integrating over z in the second term :

D = ∫_{0}^{1} dz \frac{1}{\sqrt{p^{2} + (z − 1)^{2}}} + ∫_{0}^{1} (\frac{z′}{\sqrt{p^{2} + (1 − z′)^{2}}} − \frac{z′}{\sqrt{p^{2} + z′^{2}}} ) dz′

By changing variables and combining results, this can be written simply as

D = 2 ∫_{0}^{1} \frac{1 − z}{\sqrt{p^{2} + z^{2}}} dz

The integral is elementary and gives

D = 2 [sinh^{−1}(1/p) − \sqrt{1 + p^{2}} + p]

We obtain then

L = \frac{4πN^{2}a^{2}}{lc^{2}} (Δ_{1} − Δ_{2} + Δ_{3})  (0.4)

where

Δ_{1} = ∫_{0}^{2π} sinh^{−1} (\frac{1}{p}) cos ϕ′ dϕ′

Δ_{2} = ∫_{0}^{2π} \sqrt{p^{2} + 1} cos ϕ′ dϕ′

Δ_{3} = ∫_{0}^{2π} p cos ϕ′ dϕ′

Δ_{1} − Δ_{2} can be simplified as follows. An integration by parts gives:

Δ_{1} = ∫_{0}^{2π} \frac{sin ϕ′}{\sqrt{1 + (1/p)^{2}}} \frac{1}{p^{2}} \frac{∂p}{∂ϕ′} dϕ′

With the change of variable ϕ′ = 2α, Eq. (0.3) becomes

p = x sin α  (0.5)

where we have introduced for later convenience the “aspect ratio”

x = \frac{2a}{l}

Then we obtain:

Δ_{1} = ∫_{0}^{π} \frac{sin 2α}{\sqrt{1 + p^{2}}} \frac{1}{p} \frac{∂p}{∂α} dα = ∫_{0}^{π} \frac{2 cos^{2} α}{\sqrt{1 + x^{2} sin^{2} α}} dα

and:

Δ_{1} − Δ_{2} = 2 ∫_{0}^{π} \frac{1}{\sqrt{1 + p^{2}}} [cos^{2} α − (1 + p^{2}) cos 2α] dα
= 2 ∫_{0}^{π} [\sqrt{1 + p^{2}} sin^{2} α − \frac{p^{2} cos^{2} α}{\sqrt{1 + p^{2}}}] dα

Another integration by parts gives

∫_{0}^{π} \frac{p^{2} cos^{2} α}{\sqrt{1 + p^{2}}} dα = ∫_{0}^{π} \frac{x^{2} sin α cos α}{\sqrt{1 + x^{2} sin^{2} α}} sin α cos α dα
= ∫_{0}^{π} \sqrt{1 + x^{2} sin^{2} α} (sin^{2} α − cos^{2} α) dα

and leads to the simple expression:

Δ_{1} − Δ_{2} = 2 ∫_{0}^{π} cos^{2} α\sqrt{1 + x^{2} sin^{2} α} dα  (0.6)

The integral Δ_{3} is elementary:

Δ_{3} = \frac{4a}{l} ∫_{0}^{π} sin α cos(2α) dα = −\frac{8a}{3l} = −\frac{4x}{3}

Inserting these results into Eq. (0.4), we can write

L = L_{0}f = \frac{4πN^{2}A}{lc^{2}} f

where A = πa^{2} and the correction factor f is given by

f = \frac{4}{π} ∫_{0}^{π/2} cos^{2} α\sqrt{1 + x^{2} sin^{2} α} dα − \frac{4x}{3π}  (0.7)

For numerical evaluations it is better to rewrite this result as

f = \frac{4}{π} ∫_{0}^{π/2} cos^{2} α (\sqrt{1 + x^{2} sin^{2} α} − x sin α) dα
= \frac{4}{π} ∫_{0}^{π/2} \frac{cos^{2} α dα}{\sqrt{1 + x^{2} sin^{2} α} + x sin α}  (0.8)

or to express it in terms of the hypergeometric function

F (a, b, c; z) = 1 + ∑_{n=1}^{∞} \frac{(a)_{n}(b)_{n}}{n!(c)_{n}} z^{n}  (0.9)

where (a)_{n} = a(a + 1) ⋯ (a + n − 1) = Γ(a + n)/Γ(a). Since F has the integral representation

F (a, b, c; z) = \frac{Γ(c)}{Γ(a)Γ(c − b)} ∫_{0}^{1} t^{b−1}(1 − t)^{c−b−1}(1 − zt)^{−a}dt

(for c > b > 0), we see by putting t = sin^{2} α that

F (−\frac{1}{2}, \frac{1}{2}, 2; −x^{2}) = \frac{4}{π} ∫_{0}^{π/2} cos^{2} α\sqrt{1 + x^{2} sin^{2} α}dα

and therefore

f = F (−\frac{1}{2}, \frac{1}{2}, 2; −x^{2}) − \frac{4x}{3π}  (0.10)

From this and eq. (0.9) we have the expansion, valid for x < 1,

f = 1 − \frac{4x}{3π} − ∑_{n=1}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{(−x^{2})^{n}}{(2n − 1)(n + 1)}

as can also be obtained by expanding the integrand in eq. (0.7). Explicitly, the leading terms in this series give:

f = 1 − \frac{8}{3π} (\frac{a}{l}) + \frac{1}{2} (\frac{a}{l})^{2} − \frac{1}{4} (\frac{a}{l})^{4} + \frac{5}{16} (\frac{a}{l})^{6} − \frac{35}{64} (\frac{a}{l})^{8} + ⋯  (0.11)

The theory of the hypergeometric function also provides an expansion for x > 1, given by Abramowitz and Stegun in eq. (15.3.14). Using ψ(\frac{1}{2} − n) = ψ(n + \frac{1}{2}), the result can be written

f = \frac{2}{πx} ∑_{n=0}^{∞} \binom{n − \frac{1}{2}}{n}^{2} \frac{(−1)^{n+1}}{(n + 1)(2n − 1)x^{2n}} [ln 4x − C_{n} + \frac{4n + 1}{(2n + 2)(2n − 1)}]

The leading terms of this expansion are (Grover has a mistake):

f = \frac{l}{πa} [(ln \frac{8a}{l} − \frac{1}{2}) + \frac{1}{2} (\frac{l}{4a})^{2} (ln \frac{8a}{l} + \frac{1}{4}) − \frac{1}{4} (\frac{l}{4a})^{4} (ln \frac{8a}{l} − \frac{2}{3})

+\frac{5}{16} (\frac{l}{4a})^{6} (ln \frac{8a}{l} − \frac{109}{120}) + ⋯]  (0.12)

Thus the inductance of a short coil is approximately

L = \frac{4πN^{2}a}{c^{2}} (ln \frac{8a}{l} − \frac{1}{2})

Using the properties of the hypergeometric function, it is also possible to write

f = \sqrt{1 − k^{2}} F (\frac{5}{2}, \frac{1}{2}, 2, k^{2})  (0.13)

where

k^{2} = \frac{x^{2}}{1 + x^{2}}

Equivalently, one can write f in terms of the complete elliptic integrals K and E as

f = \frac{4}{3π} \sqrt{1 − k^{2}} [\frac{K(k) − E(k)}{k^{2}} + \frac{E(k) − k}{1 − k^{2}} ]  (0.14)

Series expansions in k and in 1 − k^{2} can be obtained from (0.13) or from (0.14). There are also interpolation formulas for 0 < k < 1, corresponding to 0 < x < ∞. With an error less than 0.4%

f ≃ \frac{1}{\sqrt{1 + x^{2}}} [1 + \frac{a_{1}x^{2}}{1 + x^{2}} − \frac{4x}{3π} \frac{1}{x + \sqrt{1 + x^{2}}} + ln(1 + x^{2}) (\frac{1}{π} + \frac{0.12}{1 + x^{2}} )]

where

a_{1} = \frac{4}{π} ln 2 − 1 − \frac{1}{3π} ≃ −.223561

We conclude by giving a graph of the exact f versus x (solid line), along with the simple approximations 1−4x/3π +x^{2}/8 and (2/πx) (ln(4x) − \frac{1}{2}) (dotted line) for comparison.

(C) Self-inductance of a finite rectangular solenoid

If a and b are the sides of the rectangular cross-section, the result is (Grover, page 70)

L = L_{0}f = \frac{4πN^{2}ab}{l} f

where f has the following expression in terms of elementary functions:

f = 1 + \frac{2}{π} (f_{1} + f_{2})

f_{1} = −\frac{a}{2l} sinh^{−1} \frac{b}{a} − \frac{b}{2l} sinh^{−1} \frac{a}{b} + \frac{1}{6l} \frac{g^{3} − a^{3} − b^{3}}{ab}

f_{2} = − arctan \frac{ab}{l\sqrt{l^{2} + g^{2}}} + \frac{l}{2b} sinh^{−1} \frac{a}{l} + \frac{l}{2a} sinh^{−1} \frac{b}{l}

−\frac{l}{2b} (1 − \frac{b^{2}}{l^{2}} ) sinh^{−1} \frac{a}{\sqrt{l^{2} + b^{2}}} − \frac{l}{2a} (1 − \frac{a^{2}}{l^{2}} ) sinh^{−1} \frac{b}{\sqrt{l^{2} + a^{2}}}

+\frac{l^{2}}{3ab} [1 + \sqrt{1 + \frac{g^{2}}{l^{2}}} (1 − \frac{g^{2}}{2l^{2}} ) − \sqrt{1 + \frac{a^{2}}{l^{2}}} (1 − \frac{a^{2}}{2l^{2}} ) − \sqrt{1 + \frac{b^{2}}{l^{2}}} (1 − \frac{b^{2}}{2l^{2}} )]

with g^{2} = a^{2} + b^{2}.

For a = b we have

f_{1} = \frac{2a}{πl} [\frac{1}{3}(\sqrt{2} − 1) − ln(\sqrt{2} + 1)] = −0.4732 \frac{a}{l}

Here is a plot of f_{1} (\sqrt{\frac{1}{x}} sinh^{−1} x + \sqrt{x} sinh^{−1} (\frac{1}{x}) − \frac{1}{3} ((\frac{1}{x^{2}} + x^{2})^{3/2} − \frac{1}{x^{3}} − x^{3})) \frac{1}{π} = (1 + ln 2 + ln \frac{1}{x}) \sqrt{x} − \frac{1}{2}x + \frac{1}{12} (\sqrt{x})^{5} + \frac{1}{3}x^{3} + O ((\sqrt{x})^{9})