the creation of a collimated beam of charge, as occurs in an electron tube where a series of permanent magnets can be used to create a beam of steady current More typically, steady currents are created using wire conductors to guide the moving charge. When an external force, such as the electric field created by a battery, is applie to an uncharged conductor, the free electrons will begin to move through the positive lattice, forming a current. Each electron moves only a short distance before colliding with the positive lattice, and if the wire is bent into a loop the resulting macroscopic current will be steady in the sense that the temporally and spatially averaged microscopic current will obey V.J=0. We note from the examples above that any charges attempting to leave the surface of the wire are drawn back by the electrostatic force produced by th esulting imbalance in electrical charge. For conductors, the drift "velocity associated with the moving electrons is proportional to the applied field ud=-HeE where ue is the electron mobility. The mobility of copper (3.2 x 10-'m2/V. s) is such that an applied field of 1 V/m results in a drift velocity of only a third of a centimeter per second Integral properties of a steady current. Steady currents obey several useful inte- gral properties. To develop these properties we need an integral identity. Let f(r) and g(r)be scalar functions, continuous and with continuous derivatives in a volume region V. Let J represent a steady current field of finite extent, completely contained within V. We begin by using(B 42) to expand v·(fgJ=fg(V·J)+J.V(fg) Noting that V.J=0 and using(B 41), we get V·(fgJ)=(fJ)·Vg+(gJ)·Vf Now let us integrate over V and employ the divergence theorem fg)J·d (fJ)·Vg+(gJ)·VfdV. Since J is contained entirely within S, we must have f J=0 everywhere on S. Hence (fJ)·Vg+(gJ·Vfdv=0. We can obtain a useful relation by letting f =l and g=x; in(3. 25), where(x, y, 2)= Ji(rdV=0, (326) where Ji=Jx and so on. Hence the volume integral of any rectangular component of J is zero. Similarly, letting f=g=x; we find that Ji(r)dV=0 (327) With f = xi and g=xj we obtain Lx J; (r)+xj Ji (r)dv=0 ②2001 by CRC Press LLCthe creation of a collimated beam of charge, as occurs in an electron tube where a series of permanent magnets can be used to create a beam of steady current. More typically, steady currents are created using wire conductors to guide the moving charge. When an external force, such as the electric field created by a battery, is applied to an uncharged conductor, the free electrons will begin to move through the positive lattice, forming a current. Each electron moves only a short distance before colliding with the positive lattice, and if the wire is bent into a loop the resulting macroscopic current will be steady in the sense that the temporally and spatially averaged microscopic current will obey ∇ · J = 0. We note from the examples above that any charges attempting to leave the surface of the wire are drawn back by the electrostatic force produced by the resulting imbalance in electrical charge. For conductors, the “drift” velocity associated with the moving electrons is proportional to the applied field: ud = −µeE where µe is the electron mobility. The mobility of copper (3.2 × 10−3m2/V · s)is such that an applied field of 1 V/m results in a drift velocity of only a third of a centimeter per second. Integral properties of a steady current. Steady currents obey several useful inte￾gral properties. To develop these properties we need an integral identity. Let f (r) and g(r) be scalar functions, continuous and with continuous derivatives in a volume region V. Let J represent a steady current field of finite extent, completely contained within V. We begin by using (B.42)to expand ∇ · ( f gJ) = f g(∇ · J) + J · ∇( f g). Noting that ∇ · J = 0 and using (B.41), we get ∇ · ( f gJ) = ( f J) · ∇g + (gJ) · ∇ f. Now let us integrate over V and employ the divergence theorem: S ( f g)J · dS = V [( f J) · ∇g + (gJ) · ∇ f ] dV. Since J is contained entirely within S, we must have nˆ · J = 0 everywhere on S. Hence V [( f J) · ∇g + (gJ) · ∇ f ] dV = 0. (3.25) We can obtain a useful relation by letting f = 1 and g = xi in (3.25), where (x, y,z) = (x1, x2, x3). This gives V Ji(r) dV = 0, (3.26) where J1 = Jx and so on. Hence the volume integral of any rectangular component of J is zero. Similarly, letting f = g = xi we find that V xi Ji(r) dV = 0. (3.27) With f = xi and g = x j we obtain V  xiJj(r) + x jJi(r)  dV = 0. (3.28)