Energy and Momentum
4.1 Energy of a Charge Distribution
4.1.4 Potential Energy of Currents
While for charges we could merely separate the charges to infinity to reduce their potential energy to zero, it is far from clear that stretching current loops out to infinity would in fact diminish their energy. Instead we will use another strategy to find the energy of current loops. Unlike charge, we can make currents vanish by merely stopping the motion of charges. If we now calculate how much nondissipative work we have to do in gradually building the currents to their final value, we will have the potential energy of the currents.
When a current is established in a circuit, work must be done to overcome the induced EMF caused by the changing current. This work is fully recoverable, quite unlike the resistive losses, and does not depend on the rate that the current is increased. Only its final value and the geometry of the current loop(s) are important.
Let us consider a system composed of a number n of current loops, each carrying a time dependent currentIk. Focusing our attention on thejth loop, we note that the magnetic flux Φ in thejth loop must have the form
Φj= n k=1
MjkIk (4–33)
since the magnetic field produced at a point inside the (jth loop by any current loop k (including the jth loop under consideration) is just proportional to the current in loopk. The coefficientsMjk are known as the mutual inductance of loopj and k, whileMjj ≡Lj is the self-inductance of loopj. The EMF generated around the loop is
E=−dΦj
dt =− n k=1
Mjk
dIk
dt (4–34)
In the presence of this counter-EMF, the rate that work is done pushingIj around the loop is just
dWj
dt =− EIj =Ij(t) n k=1
Mjk
dIk
dt (4–35)
The rate that the potential energy of the entire system is changed is the sum of (4–35) over allj:
dW dt =
n j=1
n k=1
IjMjk
dIk
dt (4–36)
This is usefully rewritten as half the symmetric sum dW
dt = 12
j,k
MjkIj
dIk dt +
j,k
MkjIk
dIj dt
= 12
j,k
Mjkd
dt(IjIk) (4–37)
where we have used the symmetryMjk =Mkj.9 The total work performed to es- tablish all currents against the opposition of the counter EMF is now easily found by integrating (4–37)
W =12
j
k
MjkIkIj
=12
j
ΦjIj (4–38)
As in the case of electric charges, we would like to express the energy in terms of the potential and the field. To this end, we replace Φj with
Φj=
B ·dSj =
(∇ × A) ·dSj=
Γj
A·dj (4–39) permitting us to write
W = 12
j Γj
A·Ijdj (4–40)
This expression (4–40) is easily generalized for distributed currents to yield W = 12
A·J d 3r (4–41)
Example 4.4: Calculate the self-inductance per unit length of two parallel wires, each of radiusacarrying equal currents in opposite directions (Figure 4.2).
9The symmetry of the mutual inductance under exchange of the indices is easily seen by writing the explicit expression for the contribution of thekth current to the flux in thejth loop as follows:
MjkIk=
Bjk·dSj=
(∇ ×Ajk)·dSj= Ajk·d j
=
µ0
4π
Ikd k
|r−r|
·d j=Ikµ0
4π
d k·d j
|r−r|
where Bkj and Ajk are, respectively, the magnetic induction field and vector potential due to loopkat the position of loopj. (Note that the subscripts are not coordinate indices; no implicit summation is intended.)
Figure 4.2:Two parallel wires carry equal currents in opposite directions.
Solution: We will calculate 12 A·J d 3rand equate this to 12LI2to obtain the self- inductanceL(=Mii). For convenience we take the leftmost wire to be centered on thez-axis, while the center of the right wire lies at distanceh from the origin.10
We begin by calculating the vector potentialA both interior and exterior to a wire carrying currentI in thez direction by solving∇2A =−µ0J(Equation 3–43).
SinceA is parallel toJ in the Coulomb gauge, it suffices to solve forAz. Interior to the wire, the current density is
J= Iˆk
πa2 (Ex 4.4.1)
while it vanishes forr≥a.
In cylindrical coordinates,∇2Az=−µ0Jz becomes 1
r
∂
∂r
r∂Az
∂r
=
−µ0Jz r≤a 0 r > a
(Ex 4.4.2)
Integrating the expression in (Ex 4.2.2) forr≤atwice, we obtain Az(r≤a) =−µ0Jzr2
4 +Clnr+D (Ex 4.3.3)
BecauseAzmust be finite asr→0, we must setC = 0 leaving Az(r≤a) = µ0I
4π
D−r2 a2
(Ex 4.4.4) Forr≥a, (Ex 4.4.2) implies thatAz(r≥a) =Clnr+E. We can determineC by the requirement that the exterior magnetic induction field, ∇ × A, be given by (Ex 1.10.3)
∂Az
∂r = C
r =−Bϕ=−µ0I
2πr (Ex 4.4.5)
10We point out to the reader who was tempted to try to calculate the inductance from L= dΦ/dI that the boundary of the loop implied by Φ is ill defined. This problem recurs whenever finite-size wires are used.
We conclude that
C=−µ0I
2π (Ex 4.4.6)
Hence
Az(r≥a) =µ0I 4π
E−2 lnr a
(Ex 4.4.7) As B is finite at the boundary to the wire, A must be continuous. Matching the interior solution to the exterior solution at r =a, we obtain
µ0I 4π
D−a2 a2
= µ0I 4π
E−2 lna a
(Ex 4.4.8) from which we conclude thatD−1 =E.
To summarize, for the vector potential of a long cylindrical wire, we have Az(r≥a) =µ0I
4π
D−1−lnr2 a2
(Ex 4.4.9)
Az(r≤a) = µ0I 4π
D−r2 a2
(Ex 4.4.10) Returning to our problem, we note that interior to the wire at the origin, the vector potential is the superposition of its own interior Az (Ex 4.4.10) and the exteriorAz, (Ex 4.4.9), of the second wire carrying the same current in the opposite direction. AddingAz(r2≥a,−I) toAz(r1≤a, I) we get
Az= µ0I 4π
1−r21
a2 + lnr22 a2
(Ex 4.4.11) From the geometry, r22=r12+h2−2r1hcosϕ. With this substitution, we evaluate the volume integral ofA·J over a lengthof the wire centered on thez axis:
wire 1
J·A d 3r=µ0I2 4π2a2
2π 0
a 0
1−r21
a2 + lnr21+h2−2r1hcosϕ a2
r1dr1dϕ
= µ0I2 2π2a2
a 0
π−πr21
a2 + 2πlnh−2πlna
r1dr1
=µ0I2 2πa2
a2 2 − a4
4a2 +a2lnh a
= µ0I2 4π
1
2 + 2 lnh a
(Ex 4.4.12) This result is just twice the energy integral (4–41) over one of the wires (or equal the integral over both). Equating the energy of the currents as expressed by the preceding integral to 12LI2, we find the inductance of the wires to be
L= µ0
4π [1 + 4 ln(h/a)] (Ex 4.4.13)
It is worth noting that for ha, the logarithmic term of (Ex 4.4.13) dominates.
The constant term, µ0/4π, represents the self inductance of the current in each wire on itself.
As we previously did for electric charge distributions, we proceed now to express the potential energy of the current distribution in terms of the fields alone. Assum- ing there are no time-varying electric fields to contend with, we replaceJin (4–41) by (∇ × B)/µ0 to obtain
W = 1 2µ0
(∇ × B) ·Ad 3r (4–42)
The argument of the integral in (4–42) may be rearranged with the aid of (8) as (∇ × B)·A =B ·(∇ × A) −∇ · (A×B) = B2−∇ · (A×B), so that invoking the divergence theorem, the work expended to produce the field may be expressed as
W = 1 2µ0
B2d3r− 1
2µ0 (A×B)·dS (4–43) B2is positive definite, implying that the volume integral ofB2can only increase as the region of integration expands. By contrast, (A ×B) from a dipole field decreases asR−5 or faster, meaning that the surface integral goes to zero as 1/R3 or faster asR→ ∞. If, therefore, we include all space in the integral, we may write
W = 1 2µ0
spaceall
B2d3r (4–44)