Examples of magnetic field calculations and applications. 1 Example of a magnetic moment calculation


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1 Examples of magnetic field calculations and applications Lecture 12 1 Example of a magnetic moment calculation We consider the vector potential and magnetic field due to the magnetic moment created by a rotating surface charge, σ, on a cylinder. The geometry is shown in Figure 1. The magnetic moment, I(area), of a small loop at the position, z, as shown in the figure is ; d m = σ(rdθ dz) (πr 2 )ẑ = πσr 3 ω dz ẑ The vector potential is then; d A = µ 0 4π d m r r 3 = µ 0 4π [πσr3 ω ẑ r r 3 This is integrated using r = α z to get A. A = L/2 L/2 d A Choose α to lie in the (x, y) plane along the ˆx direction. Then ẑ r = αŷ A = µ 0 4π d m r r 3 = [ µ 0 4π [πσr3 ωα] ] dz L/2 L/2 dz 1 [α 2 + z 2 ] 3/2]ŷ The vector potential above has been calculated in the magnetic moment approximation, (ie in the first nonzero order expansion of the vector potential in a power series of the loop radius divided by the distance to the field point.) 2 The field and action of a Quadrupole lens The quadrupole field is illustrated by the magnetic field shown in Figure 2 and given by the equations; z = gx x = gz y = 0 1
2 P r z ω L/2 α dm R σ y x L/2 Figure 1: The vector potential of a rotating, cylindrical charge distribution z N S x S N Figure 2: A quadrupole magnet used to focus charged particles 2
3 The field is generated by wire windings that create the magnetic poles shown in the figure and parallel to the equipotential curves perpendicular to the field lines. We suppose the length of the field into the page is L, and the field lines are shown in the figure. The field strength increases linearly with distance from the axis. 2 = 2 x + 2 z = g 2 r 2 For positive particles moving with velocity V into the page, the Lorentz force converges a particle beam horizontally, and diverges it vertically. Check the force direction and note the further away from the axis, the stronger the force. A magnetic lens is created if two quadupole fields are placed in line and rotated by 90 deg with respect to each other. Then with a proper choice of spacing, and field strength can provide focusing of a parallel beam of particles to a point some distance behind the two magnets. 3 Power and the magnetic field The Lorentz force on a charge is F = q[ E + V ]. This force causes the charge to move in a direction perpendicular to the field and velocity, ˆl. Then we determine the power due to this movement. P = V F = q V E As indicated, the force term involving the magnetic field, V ( V ), vanishes, so the magnetic field does no work on a charge and cannot change its energy only the direction of its velocity. 4 Motion of a charged particle in a magnetic field We suppose a constant magnetic field in the ẑ direction, and a charged particle of mass, m, charge, q, and velocity, V moves in this field. The motion is given by the Lorentz force, which in nonrelativistic form, is given by the kinetic equation, F = m a. m d V = q[ V ] Neglect the interaction of the charge with other charges which may be present in a beam of such particles. In Cartesian coordinates the coupled equation set below can be produced for = ẑ. 3
4 dv x dv y dv z = V y q m = V x q m = 0 The last equation requires that V z = constant. The first two equations may be decoupled giving a second order ode. d 2 V i 2 = q m V i In the above, V i represents V x or V y. The solution is harmonic, and to satisfy both first order equations; V x = V 0x cos(ωt + φ) V y = V 0x sin(ωt + φ) Where ω = q m and φ is a phase angle to be determined from the initial conditions. Assume that at t = 0 V x = V 0x and V y = 0. Then φ = 0 and A = V 0x. Now integrate these equations again to get the coordinate trajectories. Let the initial velocity in the ẑ direction be, V z = V 0z, and the initial position for (x, y, z) is, (0, (V 0x /ω), Z 0 ). z = V 0z t + Z 0 x = (V 0x /ω) sin(ωt) y = (V 0x /ω) cos(ωt) Substitution verifies these solutions and initial conditions. We observe the motion is a helix in 3D with a projection of circular motion onto the (x, y) plane. The radius of this circle, R, is ; x 2 + y 2 = R 2 = [V 0x /ω] 2 In the above, R, represents the radius of curvature of the particle in the magnetic field. Insert the particle momentum projected onto the (x, y) plane, p. p = mv 0x = mωr = qr R = p q 4
5 5 Drift Velocity and the Lorentz force Suppose we apply a constant magnetic field in the ẑ direction and a constant electric field in the ˆx direction to a charged particle. The particle is initially at rest. The equations of motion in the (x, y) plane are; dv x dv y = q m V y + (q/m) E 0 = q m V x After application of the initial conditions, the solution has the form, with ω defined in the previous section; V x = (q/mω) sin(ωt) V y = (q/mω)[cos(ωt) E 0 /] The position is obtained by a second integration; x = (q/mω 2 )[cos(ωt)/ω] + X 0 y = (q/mω 2 )[sin(ωt)/ω (E 0 /)t] + Y 0 In the above (X 0, Y 0 ) is the initial position. This motion is strange as the circle center moves with constant velocity in the ŷ direction. Now we study this a little further by placing a resistance, proportional to the velocity, to the motion of the charge. This is artificial because force is proportional to acceleration not velocity, but you have used frictional forces proportional to velocity in mechanics, and here we want an energy disipating term. Thus add a term which contains the first odd derivative of the position, velocity. Introduction of the force term σ V into the above equations gives; dv x dv y + σ/m V x = q m V y + (q/m) E 0 + σ/m V y = q m V x Look for the equilibrum solution, ie the solution when the velocity becomes independent of time so that dv i = 0. V x = V y = qσ σ 2 + (q) 2 E 0 q 2 σ 2 + (q) 2 E 0 5
6 I V a dl b Figure 3: The geometry describing the Hall effect The drift angle between the applied field and the motion is called the Lorentz angle and is given by; tan(θ) = V y /V x = q/σ The particle drifts at a constant velocity at the angle θ with respect to the applied field. The above examples give a few simple illustrations of the motion of charged particles in magnetic fields. In general, this topic is treated in magnetohydrodynamics (plasma) for example, and the motion is highly nonlinear and nonintuitive. The understanding of plasma is crucial to the development of controlled fusion reactors. 6 The Hall effect We suppose a current through a conducting medium in which there is a magnetic field perpendicular to the current flow and the surfaces of the conductor Figure 3. The current represents a flow of charge so that there is a deflection of the current due to the Lorentz force. As previously determined Idl = qv so the force on a small element of current due to the magnetic field when the magnetic field and the current are perpendicular is; df m = Idl Charge flows and builds up an electric field on the parallel surfaces of the conductor. When equilibrum is reached, the electric force cancels the magnetic force. F = 0 = q[ E + v ] Use in the above qv = Idl. When the force reaches equlibrium; 6
7 q(de) = Idl The charge per unit volume in the conductor as obtained from the figure is ρ = q/ab(dl). Substitute for q and let de = V/b, with V the potential between the sides of the conductor. Then; V = I/(ρa) The hall effect occurs for current flow in a magnetic field and is used in a number of instruments to measure either currents or magnetic fields. 7 Example of Ampere s law and superposition There is a current flow in a cylindrical conductor of radius, R. The conductor has a hole of radius, a, displaced a distance, b from the axis of the conductor. The geometry is shown in Figure 4. It is assumed that the current density is constant in the cylinder. We are to find the magnetic field in the hole. This is done by superposition of a current in a cylinder of radius a centered at z = b in opposition to the current in the cylinder, as shown in the figure. The current in the cylinder without the hole is; I T = JπR 2 The current in the hole flowing in the opposite direction is; I H = Jπa 2 Thus the total current in the cylinder with the hole is I T I a = I C I C = J(π[R 2 a 2 ]) J = I C π[r 2 a 2 ] Now find the field at point P in Figure 4 by superimposing the contributions to the field from each of these current densities. This can be done by Ampere s law. d l = µ 0 4π I For the conducting cylinder without the hole, the enclosed current is; I c = J c (πr 2 ). The circulation on the left side of the above integral is evaluated to be (2πr). 7
8 z ^ ^ φ φ z θ P r b θ r b a R Figure 4: The geometry to find the magnetic field inside a cylindrical cavity in a cylindrical conductor C = µ 0 I πr 2 π(r 2 a 2 ) 2πr = µ 0 Ir 2π(R 2 a 2 ) In the same way the field for the current creating the hole is; H = µ 0 Ir 2π(R 2 a 2 ) The geometry is shown in the figure. Subtract these vector fields. Write the unit vectors dzẑ = r d l and use the geometry to obtain z z = b. The final result is; = µ 0 Ib 2π(R 2 a 2 ) Thus the magnetic field is constant in the interior and pointed in the ẑ direction. 8 Magnetic pressure and energy Consider two parallel current sheets separated by a distance, d, with uniform, constant currents flowing in opposite directions, Figure 5. Find the magnetic field due to one of these sheets on the other. The magnetic field is obtained using Ampere s law as previously. ecause of symmetry the magnetic field must be directed parallel to the sheet, and can only depend on the perpendicular distance from the sheet to the field point. Evaluation of the integral form of Ampere s law gives; d l = 2L = µ0 I 8
9 L I/L d I/L Figure 5: The magnetic field created by 2 parallel current sheets with currents flowing in opposi te directions The factor of 2 comes from the field above and below the sheet, with L the distance parallel to the sheet of the path along. I is the current that flows through this Amperian loop. Thus for one sheet; = µ(i/l)/2 = (µ/2) I In the above we have written I as the current per unit wih on the sheet. The direction of the magnetic field is given by the righthandrule. Note that the field is independent of the distance from the sheet. Thus the fields when superimposed from the two sheets, add between the sheets and cancel outside the sheets. Finally we also see that the force generated by the magnetic field on one sheet interacting with the current on the other is repulsive. We visualize this situation by thinking of the magnetic field as creating a pressure between the plates tending to expand the distance between them. Use the Lorentz force to calculate this force. In the equation for the Lorentz force, substitute IL for qv. The field and current direction are perpendicular, so F 2 = I 2 L 1 = I 2 x 2 (µ/2)i 1. Now the total magnetic field between the plates comes from the superposition of both fields which add to T = 2(µ/2)I 1,2. Then I 1,2 = L/µ which we substitute for I 1 in the force equation yielding; F Lx = 2µ The above is the force per unit surface area (pressure) of one current sheet on the other. This pressure attempts to push the plates apart. Now suppose we do work against this pressure by compressing the plates a distance, d. The movment of the plates removes a volume of the magnetic field, Lxd, under the Amperian loop. and puts an energy into the system given by W = Fd. We remove the geometry in the equation by dividing by the volume to obtain the energy per unit volume which we assign to the magnetic field. If we allow the plates to 9
10 d l = µ0 I T I Figure 6: The geometry to find field enclosed in a torus using Ampere s law reexpand, the plates do work removing this energy creating additional field in this volume. Compare this energy density, (1/2µ 0 ) 2, to the energy density of the electric field, (ǫ 0 /2)E 2. 9 Additional Examples 9.1 Field of a torus The geometry of a torus is illustrated in Figure 6. We assume that a current sheet moving in a circular direction around the donught is continuous, (ie very tightly wound wires). No field leaks out of the donught, and the geometry is symmetric in azimuthal angle. Thus we can use Ampere s law to get the magnetic field. Use a circular Amperian loop as shown; = µ 0 2πr In this expression I T is the total current flowing. We could write a current per unit wih of I = I T 2πr to remove the geometric dependence. 10
11 z I dz r R1 x z R2 I y Figure 7: The geometry to find the vector potential created by two long, parallel filaments of current flowing in opposite directions 9.2 Field of a bipolar filament We assume two long filaments carring a current I in opposite directions. We are to find the vector potential, A, for this geometry, Figure 7. y symmetry, A, must be independent of z. We have; A = µ 0 4π dτ Jr The current density is in the direction of ẑ and use J d area. A = µ 0 4π ẑ [ dz (R2 2 + z2 ) 1/2 A = 2 µ 0 4π I ln[z + z 2 + R2 2 ] z + z 2 + R1 2 0 ẑ A = 2 µ 0 4π I ln[r 1 R 2 ] dz (R z2 ) 1/2] 11
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