Static Electric and Magnetic Fields in Vacuum
1.2 Moving Charges
1.2.3 The Law of Biot and Savart
The magnetic induction field, B, of the long straight wire (1–32), must in fact be the sum of contributions from all parts of wire2 stretching from−∞to +∞. Since the magnetic force is related to the electric force by a Lorentz transformation that does not involve r, E andB must have the samer dependence, (1/r2). The field, d B, generated by a short segment of wire,dcarrying currentI2at the origin, must therefore be given by
d B= µ0 4π
I2d×r
r3 (1–33)
(The choice of numeric factor [µ0/4π] will be confirmed below.) Equation (1–33) is easily generalized for current segments located at r, rather than at the origin, giving
d B(r) = µ0 4π
I2d×(r−r)
|r−r|3 (1–34)
Integrating over the length of the current source, we obtain the Biot-Savart law:
B( r) = µ0 4π
I2d×(r−r)
|r−r|3 (1–35)
We might verify that this expression does indeed give the field (1–32) of the infinite straight thin wire. Without loss of generality we may pick our coordinate system as in Figure 1.8, with the wire lying along the z-axis and the field point in the (x-y plane. Thenr−r=rˆr−zˆk, |r−r|=√
r2+z2, andd = ˆk dz. The flux density,B, may now be calculated:
B( r) =µ0I2 4π
+∞
−∞
ˆk×(rˆr−zkˆ)
(r2+z2)3/2 dz= µ0I2 4π
∞
−∞
r(ˆk׈r) (r2+z2)3/2dz
=µ0I2(ˆk×r)rˆ 4π
z
√r2+z2 ∞
−∞
= µ0 2π
I2×r
r2 (1–36)
Noting that (1–36) reproduces (1–32), we consider the factorµ0/4πconfirmed.
As we will normally deal with currents that have finite spatial extent, the current elementI2din (1–35) should in general be replaced by
SJ·dSd, whereS is the cross sectionI2 occupies. The Biot-Savart law may then be written
B( r) = µ0 4π
J(r)×(r−r)
|r−r|3 d3r (1–37) Equation (1–37) plays the same role for magnetic fields as Coulomb’s law (1–7) does for electric fields.
Example 1.8: A circular loop of radiusacarrying currentI lies in thex-yplane with its center at the origin (Figure 1.9). Find the magnetic induction field at a point on thez-axis.
Solution: In cylindrical coordinates,J(r) =Iδ(r−a)δ(z) ˆϕandr−r=zˆk−aˆr.
The numerator of the integrand of (1–37) then becomes
J×(r−r) =I δ(r−a)δ(z) (zˆr+aˆk) (Ex 1.8.1) Thus (1–37) becomes:
B( 0, 0, z) = µ0 4π
I δ(r−a)δ(z) (zˆr+aˆk)
(a2+z2)3/2 rdrdϕdz
= µ0 4π
2π
0
I(−zˆr+aˆk)
(a2+z2)3/2 a dϕ= µ0 4π
2πa2Iˆk
(a2 + z2)3/2 (Ex 1.8.2) where we have used the fact that
ˆ
rdϕ= 0. In terms of the magnetic moment,
m=Iπa2k, (defined in Chapter 2, usually just current times area of the loop) ofˆ the loop, we may approximate this result at large distances as
B(0, 0, z) = 2µ0 4π
m
R3 (Ex 1.8.3)
Figure 1.9: A circular loop carrying currentIin thex-yplane.
Figure 1.10: When the current is integrated around the loop, either the limits of the integral, or the vector integrand determines the direction, in other words, the second integral in Ex 1.9.2 should usea
−a−ˆıdxas shown or alternatively−a
a ˆıdx.
The magnetic induction field of the circular current loop is of recurrent importance both experimentally and theoretically.
Example 1.9: Find the magnetic induction field at a point z above the center of a square current loop of side 2a lying in thex-yplane (Figure 1.10).
Solution: The field may be written as B(0, 0, z) = µ0
4π
J(r)×(r −r)
|r−r|3 d3r = µ0 4π
I d × (r−r)
|r−r|3 (Ex 1.9.1)
= µ0I 4π
a
−a
ˆıdx×(zkˆ−xˆı+aˆ) (z2+a2+x2)3/2 +
a
−a
−ˆıdx×(zkˆ−xˆı−aˆ) (z2+a2+x2)3/2
+ a
−a
ˆ
dy×(zkˆ−aˆı−yˆ) (z2+a2+y2)3/2 +
a
−a−dyˆ ×(zkˆ+aˆı−yˆ) (z2+a2 +y2)3/2
(Ex 1.9.2)
= µ0I 4π ·4
a
−a
aˆk dx
(z2+a2+x2)3/2 = µ0Ia2 π
2ˆk (z2+a2)√
z2+ 2a2 (Ex 1.9.3) ReplacingI(2a)2kˆbym, we recover the expression for the circular loop (Ex1.8.3) at sufficiently largez.
We have observed that a currentIdlocated atris subject to a magnetic force d F =Id×B( r). This result is easily generalized to the expression for the force between two current loops, Γ1 and Γ2 to yield the force law we alluded to at the beginning of Section 1.2.2:
F1= µ0
4π Γ1 I1d1(r1)×
Γ2
I2d2 × (r1−r2)
|r1−r2|3
(1–38)
While this expression gives a phenomenological description for calculating the force between two current carrying conductors, it is less than satisfactory in that it reimports action at a distance when it eliminatesB.
Even more than was the case for Coulomb’s law, it will be evident that using the Biot-Savart law is rather cumbersome for any but the simplest of problems. Where the geometry of the problem presents symmetries, it is frequently far easier to use Amp`ere’s law in its integral form to find the magnetic flux density. Amp`ere’s law states that
Γ
B ·d=µ0
S
J·dS (1–39)
where Γ is a curve enclosing the surfaceS.
Rather than attempting to integrate (1–37) directly, we will first obtain the differential form of Amp`ere’s law,∇ × B =µ0J, which may be integrated with the help of Stokes’ theorem to obtain (1–39).
We begin by recasting (1–37) into a slightly different form by noting a frequently used relation∇| r−r|−1=−(r−r)/|r−r|3. Making this substitution in (1–37) we obtain:
B( r) =−µ0 4π
J(r)×∇ 1
|r−r|
d3r (1–40)
We then use the identity (6) for∇× (f J) to “integrate (1–40) by parts”. Specifically,
−J( r)×∇
r−1r
=∇ ×
J(r) r−r
− 1
r−r∇ × J( r) (1–41) Because∇ acts only on the unprimed coordinates, the term∇ × J( r) vanishes and because the integral is over the primed coordinates, we can take the curl outside the integral to yield
B( r) = µ0 4π∇ ×
J( r)
r−rd3r (1–42)
As the divergence of any curl vanishes (11) we note in passing that
∇ · B( r) = 0, (1–43)
an important result that is the analogy of Gauss’ law for Electric fields. Equation (1–
43) implies via Gauss’ theorem that no isolated magnetic monopoles exist because no volume can be found such that the integrated magnetic charge density is nonzero.
We take the curl ofB as given by (1–42) to get
∇ × B( r) = µ0
4π∇ × ∇ ×
J(r)
r−rd3r (1–44) 1.2.4 Amp`ere’s Law
now using (13), we can replace∇ × ∇× by grad div − ∇2 to obtain
∇× B = µ0 4π
∇
J(r)·∇
r−1r
d3r−
J(r)∇2
r−1r
d3r
(1–45)
where we have used (7) to recast∇ ·(J /R) as J·∇(1/R). We will show the first of the integrals in (1–45) to vanish whereas the second gives, using (26),
−µ0 4π
J(r)∇2
r−1r
d3r=µ0
J(r)δ(r−r)d3r =µ0J(r) (1–46)
We return now to the first integral in (1–45). We observe that ∇f (r−r) =
−∇f(r−r) and invoke identity (7) to transform the integral to one part which can be integrated with the divergence theorem and a second that has the divergence acting onJ(r) :
µ0 4π
J(r)·∇
r−1r
d3r=−µ0 4π
J(r)·∇
r−1r
d3r
=−µ0 4π
∇·
J(r) r−r
d3r−
∇·J(r) r−r d3r
(1–47) In the case of static charge densities, as we are discussing here, the continuity equation (1–24) states that∇·J( r) = 0. Hence the second integral vanishes. We apply the divergence theorem (20) to the first integral to get
−µ0 4π
∇·
J(r) r−r
d3r=−µ0 4π
J( r)
r−r·dS (1–48) Now, the volume of integration was supposed to contain all currents so that no current crosses the boundary to the volume meaning that (1–48) also vanishes. If we wish to have non-zero current densities in the problem extending over all space, we take our volume of integration over all space. It suffices that J diminish as
|r−r|α with α < −1 or faster as r → ∞ to make the integral vanish. Having disposed of the integrals of (1–47), the integral (1–45) reduces to
∇ × B( r) =µ0J(r) (1–49) With the aid of Stokes’ theorem (18), we convert (1–49) to a line integral and obtain
Γ
B ·d=µ0
S
J·dS (1–50)
In words, Amp`ere’s law asserts that the line integral of B along the perimeter of any area equalsµ0times the current crossing that area.
Example 1.10: Find the magnetic induction field outside a long straight wire.
Figure 1.11: A circle of radiusr, centered on the wire is used to calculate Bϕnearby.
Solution: From symmetry we expect the field to be independent ofϕor z. Theϕ component of the field is then easily evaluated by integratingBϕaround a circle of radiusr centered on the wire as in Figure 1.11. Taking Bϕ to be constant on the circle, the contour integration is easily carried out:
Bϕd=µ0
A
J·dS (Ex 1.10.1)
2πrBϕ=µ0I (Ex 1.10.2)
leading to
Bϕ= µ0I
2πr (Ex 1.10.3)
Example 1.11: Find the magnetic field inside a long (assume infinite) closely wound solenoid withN with turn per unit length, each carrying the currentI. Neglect the pitch of the windings.
Figure 1.12:A currentNI threads the rectangular loop of widthWshown.
Solution: We construct a rectangular loop as shown in Figure 1.12. The inside segment lies inside the solenoid, whereas the segment completely outside the loop is placed sufficiently far from the solenoid that any field vanishes. Any supposed
radial component of the field must be the same on both sides of the loop and, its contribution therefore cancels from the integral. Again applying Amp`ere’s law,
B ·d=µ0
J·dS (Ex 1.11.1)
BzW =µ0N I (Ex 1.11.2)
We conclude that for a unit length of solenoid (W = 1),Bz =µ0N I.
If we move the segment of the rectangular loop inside the solenoid to the outside, we have zero current threading the loop, which leads to the conclusion that the field has noz-component outside the solenoid. The field inside the solenoid is evidently uniform, as the placement of the loop’s inner segment does not change our result. We have not proved the radial component of the field zero; however, since the magnetic induction field is perpendicular to the current, we conclude thatBϕvanishes whence
∇ · B = 0 requires Br to vanish.
To gain some insight into the effect of nonzero pitch of the windings, we draw an Amp`erian loop around the coil in a plane perpendicular to the coil axis. When the loop is outside the coil, exactly I crosses the plane of the loop, leading us to conclude that the field outside no longer vanishes, instead, appealing to the rotational invariance, we deduce a field corresponding to a line currentI along the axis. Inside the coil, no current crosses the loop, and the nonzero pitch has only minimal effect.
Example 1.12: Find the magnetic induction field inside a closely wound torus having a total ofN turns. Neglect the pitch of the windings.
Figure 1.13: A circle drawn in the interior of the torus includes a current NI, whereas outside the torus the included (net) current is zero.
Solution: Assuming the torus has its midplane coinciding with the x-y plane, we construct a circle parallel to thex-y plane in the interior of the torus as shown in Figure 1.15. Integrating the azimuthal component of the fieldBϕaround the circle, we obtain 2πrBϕ =µ0N I, from which we conclude that
Bϕ=µ0N I
2πr (Ex 1.12.1)
Interestingly, so long as the loop remains within the torus, up or down displacement makes no difference. The field is independent ofz but decreases radially as 1/r.
It would be convenient if B could generally be derived from a scalar potential, much in the fashion of the electric field. Unfortunately, since the curl of B is not generally zero, B cannot generally be expressed as the gradient of a scalar function. (Notwithstanding this caution, wherever the current density vanishes,B can usefully be found as the gradient of a scalar function we will call the scalar magnetic potential,Vm.)
In fact, since, according to (1–43) and to the chagrin of many physicists whose theories predict the existence, there are no magnetic charges (monopoles) analogous to electric charges,B has zero divergence.
A field without divergence (no sources or sinks) can always be expressed as the curl of a vector. We define themagnetic vector potential,A( r), by
B( r) =∇ × A( r) (1–51)
In much the same manner as the derivation of the electrical scalar potential where we expressed E as the gradient of a function, we now seek to expressB as the curl of a vector field. Glancing back at (1–42), we findB expressed in exactly this form. We conclude that
A( r)≡ µ0 4π
J( r)
|r−r|d3r (1–52)
serves as an expression for the magnetic vector potential
For many years it was thought that the vector potentialA was merely a math- ematical construct since all forces and apparently physical effects depend onB. In fact, the phase of a charged particle’s wave function depends on the line integral of the vector potential along the particle’s path. TheBohm-Aharanov experiment5 has directly demonstrated a shift in the fringe pattern of electron waves diffracted by a long magnetized needle that, though presenting no appreciable magnetic flux density along the path, does have a nonzero vector potential outside the needle.
Since forces play no central role in quantum mechanics, the disappearance of E and B from the formulation should come as no surprise. As we will see when dis- cussing the covariant formulation in Chapter 11, the potentials V and A appear to be somewhat more fundamental thanE andB in the relativistic formulation as well.
It is worth noting that according to (1–52), A is always parallel to the mean weighted current flow. We note also that from its definition, A is not unique. We can add the gradient of any function, say Λ(r), toA without changing curlA.
5See for example R. P. Feynman, R. B. Leighton, and M. Sands (1964)The Feynman Lectures on Physics, Vol. 2, Addison-Wesley Publishing Company, Reading (Mass.).