Examples of magnetic field calculations and applications 1 Example ...

Examples of magnetic field calculations andapplicationsLecture 121 Example of a magnetic moment calculationWe consider the vector potential and magnetic field due to the magnetic moment created bya rotating surface charge, σ, on a cylinder. the geometry is shown in figure 1. The magneticmoment, I(area), of a small loop at the position, z as shown in the figure is ;d⃗m =σ(Rdθ dz)dt(πR 2 )ẑ = πσR 3 ω dz ẑThe vector potential is then;d ⃗ A = µ 04πd⃗m × ⃗rr 3 = µ 04π [πσR3 ω ẑ × ⃗rr 3This is integrated using ⃗r = ⃗α − ⃗z to get ⃗ A.⃗A =L/2∫−L/2d ⃗ AChoose ⃗α to lie in the (x, y) plane along the ˆx direction. Then ẑ × ⃗r = αŷ] dz⃗A = µ 04πd⃗m × ⃗rr 3 = [ µ 04π [πσR3 ωα]L/2∫−L/2dz 1[α 2 + z 2 ] 3/2]ŷ2 The field and action of a Quadrupole lensThe quadrupole field is illustrated by the magnetic field shown in figure 2 and given by theequations;B z = gxB x = gzB y = 0The field is generated by wire windings that create the magnetic poles shown in thefigure and parallel to the equipotential curves perpendicular to the field lines. We supposethe length of the field into the page is L, and the field lines are shown in the figure. The1

PrzωL/2αdmRσyx−L/2Figure 1: The vector potential of a rotating, cylindrical charge distributionzNSxSNFigure 2: A quadrupole magnet used to focus charged particles2

field strength increases linearly with distance from the axis.B 2 = B 2 x + B 2 z = g 2 r 2For positive particles moving with velocity ⃗ V into the page, the Lorentz force convergesa particle beam horizontally, and diverges it vertically. Check the force direction and notethe further away from the axis, the stronger the force. A magnetic lens is created if twoquadupole fields are placed in line and rotated by 90 deg with respect to each other. Thena proper choice of spacing, and field strength can provide focusing of a parallel beam ofparticles to a point some distance behind the two magnets.3 Power and the magnetic fieldThe Lorentz force on a charge is ⃗ F = q[ ⃗ E + ⃗ V × ⃗ B]. This force causes the charge to movein a direction perpendicular to the field and velocity, ˆl. Then we determine the power dueto this movement.P = ⃗ V · ⃗F = q ⃗ V · ⃗EAs indicated, the force term involving the magnetic field vanishes, so the magnetic fielddoes no work on a charge and cannot change its energy.4 Motion of a charged particle in a magnetic fieldWe 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 non-relativistic form, is given by;m d⃗ Vdt= q[ ⃗ V × ⃗ B]We neglect the interaction of the charge with other charges which may be present in abeam of such particles. In Cartesian coordinates the coupled equation set below is produced.dV xdtdV ydtdV zdt= V yqBm= −V xqBm= 03

The last equation requires that V z = constant. The first two equations may be decoupledgiving a second order ode.d 2 V idt 2= − qB m V iIn the above, V i represents V x or V y . The solution is harmonic, and to satisfy both firstorder equations;V x = V 0x cos(ωt + φ)V y = −V 0x sin(ωt + φ)Where ω = qB 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 theseequations again to get the coordinate trajectories. Let the initial velocity in the ẑ directionbe, V z = V 0z , and initial position be, (0, (V 0x /ω), Z 0 ).z = V 0z t + Z 0x = (V 0x /ω) sin(ωt)y = (V 0x /ω) cos(ωt)Substitution verifies these solutions and initial conditions. We observe the motion is ahelix in 3-D with a projection of circular motion onto the (x, y) plane. The radius of thiscircle, R, is ;x 2 + y 2 = R 2 = [V 0x /ω] 2In 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 = qBRR = pqB5 Drift Velocity and the Lorentz forceSuppose we apply a constant magnetic field in the ẑ direction and a constant electric fieldin the ˆx direction to a charged particle. The particle is initially at rest. The equations ofmotion in the (x, y) plane are;4

dV xdtdV ydt= qB m V y + (q/m) E 0= − qB m V xAfter application of the initial conditions, the solution has the form, with ω defined inthe previous section;V x = (q/mω) sin(ωt)V y = (q/mω)[cos(ωt) − 1]The position is obtained by a second integration;x = −(q/mω)[cos(ωt)/ω − 1] + X 0y = (q/mω)[sin(ωt)/ω − t] + Y 0In the above (X 0 , Y 0 ) is the initial position. This motion is strange as the circle centermoves with constant velocity in the ŷ direction. Now we study this a little further by placinga resistance, proportional to the velocity, to the motion of the charge. This is artificialbecause force is proportional to acceleration not velocity, but you have used frictional forcesproportional to velocity in mechanics, and here I want an energy disipating term. Thus Iadd a term which contains the first odd derivative of the position, velocity. Introduction ofthe force term σ ⃗ V into the above equations gives;dV xdtdV ydt+ σ/m V x = qB m V y + (q/m) E 0+ σ/m V y = − qB m V xWe look for the equilibrum solution, ie the solution when the velocity becomes independentof time so dV idt = 0.V x =V y =qσσ 2 + (qB) 2 E 0q 2 Bσ 2 + (qB) 2 E 0The drift angle between the applied field and the motion is called the Lorentz angle andis given by;tan(θ) = V y /V x = −qB/σ5

IVadlbBFigure 3: The geometry describing the Hall effectThe 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 inmagnetic fields. In general, this topic is treated in magnetohydrodynamics (Plasmas) andthe motion is highly non-linear and non-intuitive. The understanding of plasma is crucial tothe development of controlled fusion reactors.6 The Hall effectWe suppose a current through a conducting medium in which there is a magnetic field perpendicularto the current flow and the surfaces of the conductor figure 3. The currentrepresents a flow of charge so that there is a deflection of the current due to the Lorentzforce. As previously determined Idl = qV so the force on a small element of current due tothe magnetic field when the magnetic field and the current are perpendicular is;dF m = IBdlCharge flows and builds up an electric field on the parallel surfaces of the conductor.When equilibrum is reached, the electric force cancells the magnetic force.⃗F = 0 = q[ ⃗ E + ⃗ V × ⃗ B]In this case;q(dE) = −d(V/w) IBdlThe 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.6

z^ ^φ φz’ θ Pr’b θ rbaRFigure 4: The geometry to find the magnetic field inside a cylindrical cavity in a cylindricalconductorThen;V = −IB/(ρa)The hall effect occurs for all current flow in a magnetic field and is used in a number ofinstruments to measure either currents or magnetic fields.7 Example of Ampere’s law and superpositionThere is a current flow in a cylindrical conductor of radius, R. The conductor has a holeof radius, a, displaced a distance, b from the axis of the conductor. The geometry is shownin figure 4. We are to find the magnetic field in the hole. This is done by superpositionas shown in the figure. For the solid conductor the current density is J ⃗ C = Iπ[R 2 − a 2 ]ẑ.We subtract the field due to the current density of the hole J ⃗ H =Iπ[R 2 2 Now apply− a ]ẑ.Ampere’s law to get the field of each cylindrical conductor at a field point, P, shown in figure4.∮ ⃗B · d ⃗ l =µ 04π IFor the conducting cylinder, the enclosed current is; I c = J c (πr 2 ). The circulation onthe left side of the above integral is evaluated to be B(2πr). Thus⃗B C =µ 0 I πr 2π(R 2 − a 2 ) 2πr ˆφ C =µ 0 Ir2π(R 2 − a 2 ) ˆφ C7

In the same way the field for the current creating the hole is;⃗B H =µ 0 Ir ′2π(R 2 − a 2 ) ˆφ HThe geometry is shown in the figure. We subtract these vector fields. Write the unitvectors ˆφ C and ˆφ H in Cartesian coordinates and use the geometry to obtain;ˆφ H = −sin(θ ′ )ẑ − cos(θ ′ )ˆxˆφ C = cos(θ)ẑ − sin(θ)ˆxr ′ = z ′ /cos(θ ′ ) = α/sin(θ ′ )r = z/cos(θ) = α/sin(θ)Since z − z ′ = b the final result is;B =µ 0 Ib2π(R 2 − a 2 )As an alternative use ⃗rˆφ c = ⃗r × dˆl and the equivalent expression for the primed coordinatesand subtract to get ⃗ b. Thus the magnetic field is constant in the interior and pointedin the ẑ direction.8 Magnetic pressure and energyConsider two parallel current sheets separated by a distance, d, with uniform, constant currentsflowing in opposite directions, figure 5. We find the magnetic field due to one ofthese sheets on the other. The magnetic field is obtained using Ampere’s law as previously.Because of symmetry the magnetic field must be directed parallel to the sheet, and can onlydepend on the perpendicular distance from the sheet to the field point. Evaluation of theintegral form of Ampere’s law gives;∮ ⃗B · d ⃗ l = 2BL = µ0 IThe factor of 2 coming from the field above and below the sheet, and L being the distanceparallel to the sheet over the path along ⃗ B. I is the current that flows through thisAmperian loop. Thus for one sheet;8

LI/LBBBBBdI/LFigure 5: The magnetic field created by 2 parallel current sheets with currents flowing inopposi te directionsB = µ(I/L)/2 = (µ/2) IIn the above we have written I as the current per unit width on the sheet. The directionof the magnetic field is given by the right-hand-rule. Note that the field is independent ofthe distance from the sheet. Thus the fields when superimposed from the two sheets, addbetween the sheets and cancel outside the sheets. Finally we also see that the force generatedby the magnetic field on one sheet interacting with the current on the other is repulsive. Wevisualize this situation by thinking of the magnetic field as creating a pressure between theplates tending 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 LB 1 =I 2 x 2 (µ/2)I 1 . Now the total magnetic field between the plates comes from the superpositionof both fields which add to B T = 2(µ/2)I 1,2 . Then I 1,2 = BL/µ which we substitute for I 1in the force equation yielding;FLx = 12µ 0B 2The 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 pressureby compressing the plates a distance, d. The movment of the plates removes a volumeof the magnetic field, Lxd, under the Amperian loop. and puts an energy into the systemgiven by W = Fd. We remove the geometry in the equation by dividing by the volume toobtain the energy per unit volume which we assign to the magnetic field. If we allow theplates to re-expand, the plates do work removing this energy creating additional field in thisvolume. Compare this energy density, (1/2µ 0 )B 2 , to the energy density of the electric field,(ǫ 0 /2)E 2 .9

∮ ⃗B · d ⃗ l = µ0 I TIBFigure 6: The geometry to find field enclosed in a torus using Ampere’s law9 Additional Examples9.1 Field of a torusThe geometry of a torus is illustrated in figure 6. We assume that a current sheet movingin a circular direction around the donught is continuous, (ie very tightly wound wires). Nofield leaks out of the donught, and the geometry is symmetric in azimuthal angle. Thus wecan use Ampere’s law to get the magnetic field. Use a circular Amperian loop as shown;B = µ 02πrIn this expression I T is the total current flowing. We could write a current per unitwidth of I = I T2πrto remove the geometric dependence.9.2 Field of a bipolar filamentWe assume two long filaments carring a current I in opposite directions. We are to find thevector potential, ⃗ A, for this geometry, figure 7. By symmetry, ⃗ A, must be independent ofz. We have;10

zIdzrR1xzR2IyFigure 7: The geometry to find the vector potential created by two long, parallel filamentsof current flowing in opposite directions⃗A = µ 04π∫dτ⃗ JrThe current density is in the direction of ẑ and use J ⃗ · d area ⃗ .⃗A = µ ∞∫04π ẑ [ dz∞∫−∞ (R2 2 + z2 ) 1/2 − √⃗A = 2 µ 04π I ln[z + z 2 + R22 √ ] ∞z + z 2 + R12 0 ẑ⃗A = 2 µ 04π I ln[R 1R 2]−∞dz(R 1 2 + z2 ) 1/2]11