1. (J - 6.1) Inductance

(a)

Show that for a system of current-carrying elements in empty space the total energy in the magnetic field is

W = \frac{1}{2c^{2}} ∫ d^{3}x ∫ d^{3}x′ \frac{J(x) · J(x′)}{|x − x′|}  (1.1)

where J(x) is the current density.

————————————— Solution ————————————— Start from

W = \frac{1}{8π} ∫ B^{2}(x) d^{3}x  (1.2)

and use

B^{2} = B · ∇ × A = ε_{ijk}B_{i}∂_{j}A_{k} = ε_{ijk}∂_{j} (B_{i}A_{k}) − ε_{ijk} (∂_{j}B_{i}) A_{k}

Discarding the boundary term,

∫ B^{2}(x) d^{3}x = − ∫ ε_{ijk} (∂_{j}B_{i}) A_{k} d^{3}x = ∫ (∇ × B) · A d^{3}x

Using ∇ × B =\frac{4π}{c} J−\frac{1}{c} \frac{∂E}{∂t} and neglecting the displacement current,

W = \frac{1}{2c} ∫ J(x) · A(x) d^{3}x  (1.3)

Equation (1.1) follows by using

A(x) =\frac{1}{c} ∫ \frac{J (x′)}{|x − x′|} d^{3}x′  (1.4)

which is correct for steady currents (consistently with the omission of the displacement current).

Comment :

Equations (1.1) and (1.2) are manifestly gauge invariant, but (1.3) is not. If we change gauge, we get in (1.3) the additional term

\frac{1}{2c} ∫ J(x) · ∇Λ(x) d^{3}x

However, this term vanishes when ∇ · J = 0 and J(x) is confined to a finite volume, as can be seen with an integration by parts. Recall that ∇ · J = 0 for steady currents. So, there is no problem.

————————————————————————— (b)

If the current configuration consists of n circuits carrying currents I_{1}, I_{2}, …, I_{n}, show that the energy can be expressed as

W = \frac{1}{2} ∑_{i=1}^{n} L_{i}I_{i}^{2} + ∑_{i=1}^{n} ∑_{j>1}^{n} M_{ij}I_{i}I_{j}

Exhibit integral expressions for the self-inductances (L_{i}) and the mutual inductances (M_{ij})

————————————— Solution ————————————— If the current density in the i-th circuit is J_{i}(x), the current is

I_{i} = ∫ J_{i}(x) · n da

where the integral is taken over a cross-section of the current flux tube. We can define a normalized current density j_{i}(x) = (1/I_{i})J_{i}(x). It then follows from (1.1) that

L_{i} = \frac{1}{c^{2}} ∫ d^{3}x ∫ d^{3}x′ \frac{j_{i}(x) · j_{i}(x′)}{|x − x′|}

M_{ij} = \frac{1}{c^{2}} ∫ d^{3}x ∫ d^{3}x′ \frac{j_{i}(x) · j_{j}(x′)}{|x − x′|}

Alternatively, we can first compute the normalized vector potential

a_{i} (x) = \frac{1}{c} ∫ d^{3}x′ \frac{j_{i}(x′)}{|x − x′|}

and then

L_{i} = \frac{1}{c} ∫ j_{i}(x) · a_{i}(x) d^{3}x

M_{ij} = M_{ji} = \frac{1}{c} ∫ j_{i}(x) · a_{j}(x) d^{3}x

An approximate expression for M_{ij} is obtained when the field of circuit C_{i} is approximately constant over the cross section of the elements in circuit C_{j}, and vice versa:

M_{ij} = \frac{1}{c^{2}} ∮_{C_{i}} ∮_{C_{j}} \frac{dl_{i}·dl_{j}}{|x_{i} − x_{j}|}

M_{ij} = \frac{1}{c} ∮_{C_{i}} dl_{i} · a_{j} = \frac{1}{c} ∮_{C_{j}} a_{i} · dl_{j}

or, introducing the normalized magnetic field b_{i} = ∇ × a_{i},

M_{ij} = \frac{1}{c} ∮_{S_{i}} n_{i} · b_{j} dS_{i} = \frac{1}{c} ∮_{S_{j}} b_{i} · n_{j} dS_{j}

Hence cM_{ij} is the magnetic flux emanating from C_{i} that is “linked” through C_{j}. A similar expression for L_{i} cannot be written in general because the self-inductance of an infinitely thin wire is divergent.

2. (J - 6.2) Inductance of transmission line

A two-wire transmission line consists of a pair of nonpermeable parallel wires of radii a and b separated by a distance d > a + b. A current flows down one wire and back the other. It is uniformly distributed over the cross-section of each wire. Show that the self-inductance per unit length is

c^{2}L = 1 + 2 ln(d^{2}/ab)  (2.1)

————————————— Solution ————————————— We start by computing the magnetic field generated by a circular wire of

radius a centered at the origin. Formulas are simpler to write if we assume that the integrated current is c (rather than 1 or generically I). For ρ > a we have the well-known result of Biot and Savart

B = 2/ρ

while for ρ < a, keeping in mind that the current density is c/πa^{2}, we find 2πBρ = 4πρ^{2}/a^{2} or

B = 2ρ/a^{2}

The direction of B is that of

ê_{φ} = (− sin φ, cos φ) = (−\frac{y}{ρ}, \frac{x}{ρ})

Similarly we can compute the magnetic field of the return wire that is located at (d, 0) and carries current −c. If we define

ρ′ = \sqrt{(x − d)^{2} + y^{2}} = \sqrt{ρ^{2} + d^{2} − 2dρ cos φ}

we find that the magnitude of the field is B′ = 2/ρ′ for ρ′ > b and B′ = 2ρ′/b^{2} for ρ′ < b. The direction of B′ is that of

ê′_{φ} = − (−\frac{y}{ρ′} , \frac{x − d}{ρ})

so that

ê_{φ}·ê′_{φ} = − \frac{y^{2} + x (x − d)}{ρρ′} = − \frac{ρ − d cos φ}{ρ′}

One way to find c^{2}L is to compute

\frac{1}{8π} ∫ (B + B′)^{2} d^{2}x

There are 3 pieces to the integral and the computation is lengthy.

It is more convenient to use instead the formula

c^{2}L = \frac{1}{πa^{2}} ∫_{ρ<a} (A + A′) d^{2}x − \frac{1}{πb^{2}} ∫_{ρ′<b} (A + A′) d^{2}x′  (2.2)

where A is the z component of the vector potential that gives B, and similarly for A′. Using B_{φ} = −∂A/∂ρ, we find

A = \{ \begin{matrix}− ln \frac{ρ^{2}}{a^{2}} & for ρ > a \\ 1 − \frac{ρ^{2}}{a^{2}} & for ρ < a \end{matrix}

and similarly

A′ = \{ \begin{matrix}ln \frac{ρ^{′ 2}}{b^{2}} & for ρ′ > b \\ −1 + \frac{ρ^{′ 2}}{b^{2}} & for ρ′ < b \end{matrix}

Thus

\frac{1}{πa^{2}} ∫_{ρ<a} A d^{2}x = ∫_{0}^{a} (1 − \frac{ρ^{2}}{a^{2}} ) \frac{2ρ dρ}{a^{2}} = \frac{1}{2}

\frac{1}{πa^{2}} ∫_{ρ<a} A′ d^{2}x = \frac{1}{πa^{2}} ∫_{0}^{a} ρ dρ ∫_{0}^{2π} ln \frac{ρ^{′ 2}}{b^{2}} dφ = ln \frac{d^{2}}{b^{2}}

where to compute that last integral we have used

∫_{0}^{2π} ln \frac{ρ^{′ 2}}{b^{2}} dφ = 2π ln \frac{d^{2}}{b^{2}}

which follows from

ρ^{′ 2} = ρ^{2} + d^{2} − 2dρ cos φ = (d − ρe^{iφ}) (d − ρe^{−iφ})

∫_{0}^{2π} ln (d − ρe^{±iφ}) dφ = 2π ln d+ ∫_{0}^{2π} ln (1 − \frac{ρ}{d}e^{±iφ}) dφ = 2π ln d

as can be seen by expanding the logarithm and integrating term by term. Collecting results,

\frac{1}{πa^{2}} ∫_{ρ<a} (A + A′) d^{2}x = \frac{1}{2} + ln \frac{d^{2}}{b^{2}}

and similarly

−\frac{1}{πb^{2}} ∫_{ρ<b} (A + A′) d^{2}x = \frac{1}{2} + ln \frac{d^{2}}{a^{2}}

Inserting back in (2.2) we obtain (2.1) in the form

c^{2}L = 1 + ln \frac{d^{2}}{a^{2}} + ln \frac{d^{2}}{b^{2}}

3. (J 6.3) Inductance of coaxial cable

A circuit consists of a thin long conducting shell of radius a and a parallel return wire of radius b on axis inside. If the current is assumed distributed uniformly throughout the cross section of the wire, calculate the self-inductance per unit length. What is the self-inductance if the inner conductor is a thin hollow tube?

————————————— Solution —————————————

We start by computing the magnetic field generated by a circular wire of radius a centered at the origin. Formulas are simpler to write if we assume that the integrated current is c (rather than 1 or generically I). For ρ > a we have the well-known result of Biot and Savart

B = 2/ρ

while for ρ < a, keeping in mind that the current density is c/πa^{2}, we find 2πBρ = 4πρ^{2}/a^{2} or

B = 2ρ/a^{2}

The direction of B is that of

ê_{φ} = (− sin φ, cos φ) = (−\frac{y}{ρ}, \frac{x}{ρ})

The field of the thin shell is such that the total field is B inside the volume bounded by the shell, and zero outside.

Then we simply compute the energy per unit length

\frac{1}{8π} ∫ B^{2} d^{2}x = ∫_{0}^{a} \frac{ρ^{2}}{a^{4}} ρ dρ + ∫_{a}^{b} \frac{1}{ρ^{2}} ρ dρ = \frac{1}{4} + \frac{1}{2} ln \frac{b^{2}}{a^{2}}

and equate it to \frac{1}{2}Lc^{2}. We obtain

Lc^{2} = \frac{1}{2} + ln \frac{b^{2}}{a^{2}}

If the inner conductor is a thin hollow tube, the result is Lc^{2} = ln (b^{2}/a^{2}) .

Comment :

A thin hollow tube is equivalent to a “perfect magnetic conductor” in the sense that all the current is on the surface. Then, according to problem 6.5 the capacitance per unit length C is related to L by LC = 1/c^{2}. It is well known that for the coaxial cable 1/C = ln (b^{2}/a^{2}).