15 Inductance solenoid, shorted coax

Transcription

1 z 15 nductance solenoid, shorted coax 3 Given a current conducting path C, themagneticfluxψ linking C can be expressed as a function of current circulating around C. 2 1 Ψ f the function is linear, i.e., if we have a linear flux-current relation 0 1 Ψ=L, 2 then constant x L = Ψ is termed the self-inductance 1 of path C, anelementaryinductor. Differentiating the flux-current relation with respect to time t, and using the fact that E = dψ, we find that the emf of inductor L is simply E = L d, which is a voltage rise across the inductor in the direction of current (with L d denoting the voltage drop in the same direction as used in circuit courses). 1 AmutualinductanceM 12,bycontrast,relatesthefluxlinkingcoilC 2 to a current 1 flowing in a second coil C 1. 1, E = L d + V (t) =L d -

2 For an inductor consisting of n-loops, the emf E measured around n-loops is naturally implying an inductance E = n( d Ψ) = d nψ Ld L nψ. Example 1: An n-turn coil has a resistance R =1Ωand inductance of 1 µh. f it is conducting 3 A of current at t =0,determine(t) for t>0. Solution: Current flow in the resistive n-turn coil will be driven by emf E = L d matching the voltage drop R. Hence L d = R d + R L =0 (t) =(0)e R L t =3e 106t A. As illustrated by above example, current around a resitive loop C will in general be obtained by solving a differential equation constructed using the emf of the loop. = E used last lecture assumes that emf produced by the induced R current is small compared to an externally produced emf. We continue with typical inductance calculations. 2

3 nductance of long solenoid: Consider a long solenoid with length l, cross-sectional area A, and a density of N loops per unit length as examined in Example 3 of Lecture 12 (see figure in the margin). As determined in Example 3, the magnetic flux density in the interior of the solenoid is B =ẑb B = µ o Nẑ while n = Nl is the number of turns of the solenoid. Thus, the inductance of the solenoid with n = Nl turns is L = nψ = Nl(µ on)a = N 2 µ o Al. As we know from our circuit courses, an inductor L such as the solenoid coil considered above can be used to store energy. An inductor connected to an external circuit with a quasi-static current develops a voltage drop V = L d across its terminals 2 and absorbs power at an instentaneous rate implying a stored energy of P = V = L d = d (1 2 L2 ), W = 1 2 L2 = 1 2 N 2 µ o Al 2 = B z 2 in an inductor in a conducting state. 2µ o Al = 1 2 µ o H z 2 Al 2 Assuming a physical size much smaller than a wavelength λ = c/f for the highest frequency in (t). l B = µ o N 3

4 Notice that the stored energy of the solenoid is 1 2 µ o H z 2 = 1 2 µ oh H times its volume Al occupied by the field H inside the solenoid. That suggests that w = 1 2 µ oh H can be interpreted as stored magnetostatic energy per unit volume in general. Also both inductance L and stored energies W and w would have µ replacing µ o in material media with permeabilities µ =(1+χ m )µ o and magnetic susceptibilities χ m,inanalogywiththeconceptsof permittivity ɛ =(1+χ e )ɛ o and electrical susceptibility χ e. Permeability and magnetic susceptibility notions will be examined in a future lecture. 4

5 nductance of shorted coax: Consider a coaxial cable of some length l which is shorted at one end (with a wire connecting the inner and outer conductors), so that a steady current can flow on the inner conductor of radius a to return on the interior surface of the outer conductor at radius b after having circulated through the short. Wewillnextdeterminethe inductance L of such an inductor after first computing the magnetic flux density B φ that will be produced by the inner conductor current. n B φ calculation we will assume l b so that an infinite coax approximation can be invoked. Expanding the integral form of Ampere s law B dl = µ o C as C B φ 2πr = µ o over a circular integration contour C of a radius r>a,wefinhat the magnetic flux density in the interior of the coax cable is B φ = µ o 2πr. Therefore, the magnetic flux linked by the closed current path (see figure in the margin) is Ψ= B ds = l µ b o 2π dr r = lµ o 2π ln b a. S 5 a b a B r l Short Shorted coax circulates a current linking a magnetic flux Ψ confined to a region bounded by the outer conductor of the coax. z

6 Clearly, we have a linear relation Ψ=L, with L ln b a 2π lµ o, which is the inductance of a shorted coax of a finite length l. The inductance of the coax per unit length is L = ln b a 2π µ o, which should be contrasted with capacitance per unit length of the same coax configuration. C = 2π ln b ɛ o a Notice how L and C are proportional to ɛ o and µ o,respectively, having proportionality constants which are inverses of one another. 6

7 nductance of shorted parallel plates: f a pair of parallel plates of areas A = Wl and separation d were shorted at one end, we would obtain effectively an inductor with a per length inductance L = d W µ o that accompanies per length capacitance C = W d ɛ o of the same parallel plate configuration. This follows from a generalization of our finding above that the proportionality constants of L and C are arithmetic inverses of one another. 7