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UIUC 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

LECTURE NOTES 16

 THE MAGNETIC VECTOR Ar

 We saw in that E 0 {always} (due to intrinsic / microscopic nature of the  electrostatic ) permitted us to introduce a potential Vr such that:   Er Vr  {n.b. Vr is uniquely defined, up to an (arbitrary) constant.}

 Analogously, in , the Br 0 (always) {no magnetic charges / no   magnetic monopoles} permits us to introduce a magnetic Ar such that:

   B rAr  S.I. units of the magnetic vector potential Ar = Tesla-meters      Teslas 1 m Tesla- Meters

    Then: Br Ar  0 {always}   The of a of a Fr is always zero

Ampere’s Law:      2 In differential form: BrArArArJr         o free   Now, just as in the case of electrostatics, where Vr was uniquely defined up to an arbitrary  constant Vo , then let: Vr Vr  Vo 0        then: Er V  r  Vr   Voo  Vr  V Vr   i.e. Er V  r  Vr 

An analogous thing occurs in magnetostatics - we can add / we have the freedom to add to the     magnetic vector potential Ar the of any scalar function rrm  where   2 m r magnetic SI Units of magnetic scalar potential m r Tesla-m

   Then: Ar Ar   r Ar m  r  Formally known as a Gauge Transformation

 The curl of the gradient of a (m r here) automatically/always vanishes, i.e.:  0Always  !!!             BrArArrArrArr            m     Ar

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 1 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  Note that the magnetic scalar potential m r has same physical units as magnetic m :    Tesla-m2 = (,  BrdA !!) eeek!!!! m S

  Please do NOT confuse the magnetic scalar potential m r (= a scalar point function,  whose value can change at each/every point in space, r ) with the magnetic fluxm (which is a  constant scalar quantity (i.e. a pure number), independent of position) mmr !!!

   Thus, like the scalar potential Vr, the magnetic vector potential Ar is (also) uniquely   defined, but only up to an (arbitrary) vector function rrm .

   Ar Ar   r Ar m  r

   The definition BrAr  specifies the curl of Ar, but in order to fully specify the        vector field Ar, we additionally need to specify the divergence of Ar, Ar.

    We can exploit the freedom of the definition of Ar to eliminate the divergence of Ar        - i.e. a specific choice of Ar will make Ar divergenceless: Ar 0  Gauge

   If: Ar Ar   r Ar m  r     2  Then: A r  Ar   r Ar mm  r  Ar   r

  While the original magnetic vector potential, Aris not/may not be divergenceless, we can       make Ar Ar   r Ar m  r divergenceless, i.e. Ar 0 if we chose     2 rrm  such that rrArm    Coulomb Gauge

A Simple Illustrative Example:   Suppose  a region of space that has a uniform/constant , e.g. BrBz o ˆ .    Ary  Arx  Then: BrBzAro ˆˆ    z. xy      Thus (here): Ar A  r xˆˆ A  r y A  r zArxAryˆ ˆˆ  xy z xy 0  1  1  11 If Ax rBy o and AyorBx , then Ar Byxˆˆ  Bxy, and thus we see that this 2 2 22oo choice of magnetic vector potential indeed gives us the correct B -field:    Ar y  Arx  11 BrAr    zBBBzˆˆ 22ooo   xy

2 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

  1 1   A Ay A By  Bx  0 Is A 0 satisfied? A x  z 2 o  2 o   0 Yes!!!  xyz  x y z Note that we could also have instead chosen/used a different magnetic vector potential:        Ar Ar   r Ar m  rwhere e.g. rrAmo   , i.e. where Ao is any     (arbitrary) constant vector, AAxAyAzooxoyozˆˆˆ . Since (here) rrAmo   , then    rrAAxAyAzm   o  oxˆˆ  oy  oz ˆ means that the gradient of the magnetic scalar

 Ax Ayoy  Az   potential (here) is:rxyzAxAyAzAr oxˆˆ   oz ˆˆ  ˆˆ      moxoyozoxyz  and thus the magnetic scalar potential itself (here) is: moxoyozrAxxAyyAzzˆˆ  ˆ .

  Thus, here for the case of a constant/uniform magnetic field BrBz o ˆ we see that there is    in fact a continuum of allowed magnetic vector Ar Ar  Aom Ar   r     that simultaneously satisfy BrBzArˆ   and Ar 0 with the addition of an o  (arbitrary) constant magnetic vector potential AAxAyAzooxoyozˆˆˆ contribution with  corresponding magnetic scalar potential moxoyozrAxxAyyAzzˆˆ  ˆ . Note that this is exactly  analogous to the situation in electrostatics where the scalar Vris unique, up to an arbitrary constant, Vo because there exists no absolute reference in our universe – i.e. absolute measurements of the scalar electric potential are meaningless - only potential differences have physical significance!!!

  We used this simple example of the constant/uniform magnetic field BrBz o ˆ to elucidate   this particular aspect of the magnetic vector potential Ar. Here in this particular example, we found that the addition of an arbitrary constant vector      rAAxAyAzooxoyozmˆˆ  ˆ   rto the magnetic vector potential Ar was allowed,     i.e. Ar Ar   r Ar  Ao , which leaves the magnetic field Br unchanged. In general there are many instances involving more complicated physics situations,       where Br  constant vector field, where indeed BrAr   and Ar 0 are  simultaneously satisfied for Ar Ar   r, because it is possible to determine/find a    2  corresponding magnetic scalar potential m r for the problem satisfyingm rAr  ,   but it is (very) important to understand that, in general, the allowed rrm  (very likely) may not be simply a constant vector field, but indeed one which varies in space (i.e. with   position vector, r )! Here again, however, the new Ar Ar   rwill also be such that   BrAr  will be unchanged, exactly analogous to Vr Vr  Vo leaving  Er V  runchanged.

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 3 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede     2   So we see that if m rAr   then yes, Ar 0 .

     It is always possible to find an rrm  in order to make Ar 0 .

Note however that this situation is then formally mathematically identical to Poisson’s Equation,  for the magnetic scalar potential m r because:  2  2  Tot r mmrr    Analogous to Vr in electrostatics!!!  0  Equivalent magnetic Physically, m r could e.g. be due to bound effective charge densit y  magnetic charges associated with a magnetic material…  If we assume that the equivalent magnetic volume , m r  0 and we want    Ar 0   2  Then: Ar  r  Ar m  r 0       Or: Arm  r 0  Arm  r

 Then, the solution to Poisson’s equation for the magnetic scalar potential m r is of the form:    1  r  1  r rdm    Analogous to Vr Tot d  in electrostatics m v v 4 r 4 0 r  with rrr

 (n.b. these two relations are both valid assuming that m rr and ToT vanish when r !)

    So then if m rAr , and m rAr  vanishes as r , then the magnetic  scalar potential m r is given by:

   11 rAr  rdm    d m 44vvrr

   {Note that if Arm  r does not go to zero at infinity, then we’ll have to use some other  means in order to obtain an appropriate m r , e.g. in an analogous manner to that which we’ve  had to do for the (electric) scalar potential Vrassociated with problems that have distributions extending out to infinity.}

 Thus, this choice of m r ensures that indeed we can always make the magnetic vector      potential Ardivergenceless, i.e. the conditionAr 0 (Coulomb Gauge) can always be    met, for the case of magnetostatics. Note that if m r  0 then m rAr  0 .

4 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

With the choice of the magnetic scalar potential:   1  r       rdm   and Ar Ar  A Ar   r, and Ar 0 m 4 v r om

Then Ampere’s Law (in differential form) becomes:

     B rArArArJr       2      0 free   0 2  Which gives: Ar0 Jfree  r  Vector form of Poisson’s equation for magnetostatics.

2  Arxoxfree  J  r The three separate/independent scalar forms  2  i.e.: Aryoyfree J  r  of Poisson's equation are connected by:     2  JrJ rxJryJrzˆˆ ˆ Arzozfree J  r free x free  y free  z free 

 22 2  2  n.b. in Cartesian coordinates: Ar Arxyz  xˆˆ A  r y Ar  zˆ

However, in curvilinear coordinates (i.e. spherical-polar or cylindrical coordinates)

rxyzˆˆˆsin cos sin  sin cos ˆ  e.g. spherical-polar coordinates: ˆ cos cosxˆˆ cos  sinyz sin ˆ ˆ ˆˆ sinxy  cos

Note that the unit vectors rˆ,ˆ ˆ for spherical-polar coordinates are in fact explicit functions of    the vector position, r i.e. rrrˆˆ, ˆˆ  r and ˆˆ   r and therefore rˆ,ˆ ˆ must also be   explicitly differentiated in calculating the Laplacian 2 of a vector function (here, Ar) in curvilinear (i.e. either spherical-polar and/or cylindrical) coordinates!!! This is extremely important to keep in mind, for the future…

   The safest way to calculate the Laplacian of a vector function 2 Ar in terms of curvilinear coordinates, is to NOT use curvilinear coordinates!!! Failing that, then one should use:

   2 Ar Ar  Ar   Ar    0 in the Coulomb Gauge      If m r  0 then (automatically) m rAr  0 and we can use Ardirectly.   2  Hence, if Ar0 Jfree  r (vector Poisson equation for magnetostatics),        Jrfree   then if Jr 0 as r , then Ar 0 d  where rr rr free 4 v r

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 5 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Generalizing this for a moving point charge as well as for line, surface and volume current  densities (with BrAr ), we summarize these results in the following table:

     vr    vr  rˆ Ar o q Br o q 4 free r 4 free r 2      Ir ˆ  o free     Irdfree   r Ar d Br o C C 2 4 r  4 r   o d  dr rˆ  I o  free C 4 r  I free 4 C r 2      Kr  ˆ  0 free     Krfree  r Ar da B rda o  S  2 4 r 4 S r      Jr  ˆ  o free     Jrfree  r Ar d  B rd o   v  2 4 r 4 v r

   Note that:AvIdKJ , , , , i.e. A is always parallel to the direction of motion of current, with     relative velocityv , whereas B AvIdKJ  ,, , , .

   Note also that B and A both vanish when v  0 (e.g. in the rest frame of a current (e.g. a proton or an beam)).

A Tale of Two Reference Frames:

For a pure point electric charge/point electric monopole , q we know that if it is moving   in the lab frame with speed v << c {c = speed of light in vacuum} that the magnetic field Bq r observed in the lab frame is:

  1 q rrˆˆo   Brqq Ar  2222 vEr  q  v   qv   cc440 rr  

  Thus in the lab frame where this charged particle is moving, the magnetic vector potential Arq  associated with this moving charged particle (as observed in the lab frame) has a non-zero curl.

Contrast this with the situation in the rest frame of this pure point electric charge particle, where   the magnetic field vanishes, i.e. the magnetic vector potential Arq associated with this charged particle has no curl!!!

   1 Vrt, We will find out (next semester, in P436) that: Art, in electrodynamics. ct2     connection between the Art ,  field and electric scalar potential V  rt , - they are in fact the 3 spatial & 1 temporal components of the relativistic 4-potential in electrodynamics !!!

6 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  Uses of the Magnetic Scalar Potential m r :

In certain (limited) circumstances for magnetostatics, it is actually possible to have the   magnetic field Br directly related to the (negative) gradient of a magnetic scalar potential    m r , i.e. Brrom  , in direct analogy to that for electrostatics Er Vr .   2 2  However, whileBrom  r om  r 0 is satisfied, i.e. m r 0   is Laplace’s equation for the magnetic scalar potential,m r (n.b. implying that m r  0 ),    Ampere’s law B rrJr    is not satisfied/is violated (!!!) unless omo    0Always !!!   Jr 0 everywhere in the region(s) of interest. These current-free regions must also be simply- connected. {A region D (e.g. in a plane) is connected if any two points in the region can be connected by a piecewise smooth curve lying entirely within D. A region D is a simply connected region if every closed curve in D encloses only points that are in D.}

 The use of Brrm is in fact helpful for determining the magnetic fields associated with e.g. current-carrying filamentary wires, current loops/magnetic , and e.g. the magnetic fields associated with magnetized materials/magnetized objects.

 The SI Units of the magnetic vector potential A are Tesla-meters (= magnetic field strength per unit ), which is also equal to Newtons/Ampere ( per unit current) = kg-meter/Ampere-sec2 = kg-meter/Coulomb-sec = (kg-meter/sec)/Coulomb = per unit charge, since (kg-meter/sec) are the physical units associated with momentum p  ""mv .  Thus, for the A -field: force momentum 1 Tesla-meter = 1 unit of = 1 unit of Ampere of current Coulomb of charge   and for the B -field, from BrAr : forceN momentum 1 Tesla = 1 unit of meter meter Ampere of currentAm Coulomb of charge  Physically, the A -field has units of force per Ampere of current (or momentum per Coulomb of  electric charge), and thus physically, the magnetic field BrAr  is the curl of the force per unit current (or momentum per unit charge) field. Note also that force, F  dP dt and  current, I  dQ dt such that the magnetic vector potential A physically also has units of ForceFdpdtdpdt / / p     and thus B is the curl of this . CurrentIdqdt / dq/ dt q

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 7 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

The Magnetic Vector Potential of a Long Straight Wire Carrying a Steady Current

For a filamentary wire carrying steady current I, the magnetic vector potential and magnetic field     d  rˆ  o d  o  are: Ar  I and Br Ar   I 2 4 C r 4 C r   Let the length of wire = 2L, and I  Izˆ and thus ddzdzzˆˆ

  r r rr R22 

The magnetic vector potential,   dA r Ryˆ due to the current-carrying  line segment d  carrying current I is:      oodr  dr dA r Ryˆ  I   I 44r R22 

The corresponding infinitesimal magnetic field increment is:    oodr  dzzˆ dB r Ryˆˆ  dA  r  Ry  I    I   44r R22  zL dzzˆ  L d Ar Ryˆˆ dAr   Ryoo I 2 CzL  22  0 22 44RR  Now: L oo22   22   2I lnRzˆ 2  ILRLR ln ln 440   

More generally, for a  distance R away from a long straight wire of length 2L carrying a steady current I:    L 2 Ar Ryˆ 2ln11o I R zˆ L 4 R 

  o 2L If L >> R, then: Ar Ryˆ 2ln I zˆ {Since 11 for  1.} 4 R

  Note that if L → ∞ (or R → 0), then Ar Ryˆ diverges (logarithmically)!   This is OK (unphysical anyway!) because even if A → ∞, BA !!! (necessarily!!!)

8 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

So, for a  distance R away from a long straight wire of length 2L carrying a steady current I:

  L 2 A r Ryˆˆ2ln11o I R zˆˆ  A r  Ry z L z  4 R 

 Then BrAr Let’s do this in cylindrical coordinates: (note:  x22yR here):

0 0 0 0  0 1 Az A A Az 1 A Az BR   ˆ  ˆ A  zˆ  ˆ  z z    R  R  R  R

ˆˆzˆ In Cylindrical Coordinates:  1  A Or: BR0 z ˆ zˆ Rz  R 00A z  R

 yˆ B  0  1 Az Az ˆ Thus: B   R ˆˆ R RR B  0 ˆ z xˆ 

 Az o 22 Then: BR ˆˆ R 2lnln I LLRR    4 R  

Now if: UR L L22  R  1 dU R R Then: ln UR RURdRLLR22 LR 22  dU R R  1 Since:  and: ln R  dR LR22 R R

 B-field associated  with filamentary o R 1 Then finally: BR  I  ˆ  wire of length 2L 2 22R L2 111RR carrying steady LL  current I.  o I Note that as L → ∞, then BR ˆ  i.e. exactly the same as we obtained for 2 R ∞-long straight filamentary wire carrying steady current I (see previous P435 Lecture Notes)!!!

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 9 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

   The Magnetic Vector Potential Ar and Magnetic Field BrAr  Associated with

a Pair of Long, Parallel Wires Carrying Steady Currents I1 Izˆ and I2 Izˆ , Separated by a Perpendicular Separation Distance, d

Plane  to wires and containing  separation distance d

   ddzdzz1 ˆˆ  r 111rrrr 1     ddzdzz2 ˆˆ  r 222rrrr 2 rrr12

For simplicity’s sake, assume L >> R1, R2.

  o 2L o 2L Then: AR11 Iln  zˆ and AR22 Iln  zˆ 2 R1 2 R1

Then, using the principle of linear superposition:  oo22LL   ArArArIzIzTOT 12   lnˆˆ   ln  22RR12    2  ooRR22   Or: ArTOT  Iln zˆˆ  I ln  z for L >> R1, R2. 24RR11  

10 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Now let us re-locate the local origin to be at the LHS wire, where it intersects the  -plane:

Top View

  I1 (out of page) d I2 (into page) 222 y d-y R1 xy 22 22 R1 x R2 R2 xdy 

 2 2 2  ooR2  xdy Then: ArTOT  Iln zˆˆ I ln  z for L >> R1, R2. 44Rxy  22 1 

 AyTOT  0 diverges (x = 0)     Ar Ayd 0 for I IIz  ˆ TOT  TOT 2 12

(for x = 0 i.e. observation point on yˆ -axis) y = 0 y = d/2 y = d y

 AydTOT  diverges (x = 0)

 Note that AAz ˆ i.e. AATOT TOT 0 (since currents only in zˆ -direction) !!! TOT z xy Note also that AAzTOT z ˆ changes sign – its direction is parallel to the closest current!!!

 Then: BTOTrAr TOT , in Cartesian xˆˆyzˆ coordinates:

TOT TOT Az o dy y 222 BIx  22  R1 xy y 2 RR21 TOT TOT Az o x x 22 22 BIy  22  R2 xdy  x 2 RR21 TOT Bz  0

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 11 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

At the point x = 0 and y = d/2 (i.e. at the center point, midway between the two conductors): {where R1 = R2 = R = d/2}:

dd  2 TOT  11 4 2 I BIooo22   I   I  o x 22    22dd dd 2 dd 22 22 TOT By  0 TOT Bz  0 Top View:  d 2o I BTOT xy0, , z  0  xˆ 2  d

   The Magnetic Vector Potential Ar and Magnetic Field BrAr  Associated with a Magnetic Loop (For Large Source-Observer Separation Distances)

 For simplicity’s sake, let us choose the observation / field point Pr to lie in the x-z plane:

Observation / field  Point Pr Px ,0, z  Magnetic Dipole Loop has radius R

 R  r

 I  Iˆ

12 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

 dRd  ˆ cos rrˆˆ  = opening angle between rˆ and rˆ    rr rr rr  dRd   cos    where: rrR rr rr rR {from the arc length formula “ SR  ”}

     ooIdr  dr Now: Ar  I 44CCrr 

     Ar (here) is a function of xˆˆ and y only (actually only ˆ ) and not zˆ since IdId lies in the x-y plane and I  Iˆ (here).    Since we evaluate A in the x-z plane, the only component of d that will contribute to A (there)  will be in the yˆ direction (n.b. A is parallel to the closest current from the observation point P(r)).

     We only want the component of dr along the yˆ -axis, dr cos (Note: If we wanted    to evaluate A e.g. in the y-z plane, then we would want only the component of dr along the   xˆ -axis, dr sin )

 2  o Rdcos  Then: Ar Ax ,0, z   Iˆˆ  yˆ in the x-z plane for rxz ,0, . 4 0 r  Now: r 22rR 2 2cos rR  (from the Law of Cosines) and r rr

1 2 rRrR2 212  RR  And: 122  cos   1   cos  if R  r , r and r  r . rrr2 rr   And: r r rrcos xxˆˆˆ zzˆ R cos x  R sin y   xR cos   (if the observation / field point Pr Px ,0, z  lies in the x-z plane) rRxR1cos2     Thus:  1 2 for r >> R and rrr r  r 2 rr 2   IRxR2 1  Then: Ar Ax ,0, z  o R cos  1 cos d yˆ  0  2 42 rrr   IR 2  o ˆ  Or: Ar Ax ,0, z  3 xy for r >> R and rrr r  4 r

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 13 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

But notice that in the x-z plane: zˆ  Pr Px ,0, z  x sin   r  r  z

xˆ x yˆ out of page

  IR 2  o ˆ  Ar Ax ,0, z  2 sin y for r >> R and rrr r  4 r   But in the x-z plane, ˆ  yˆ , and since the direction of Aris always parallel to the current:

   IR 2     o ˆ  Ar 2 sin for r >> R and rrr r  {n.b. Ar I ˆ } 4 r  The magnetic dipole moment associated with this current-carrying loop is mmz ˆ   mIaIazˆ (for this planar loop) where (here): aRz  2 ˆ (by the right-hand rule)   mIRz  2 ˆ and: mmIaIR22  RI Note that: zrˆˆˆcos r  sinˆˆ  r ˆ sin    r ˆ  sin   ˆˆ  sin   Thus, the quantity: mrˆˆ IazrIRˆˆ  22 zr  ˆ  IRsin ˆ

   oomrˆ  mr   Ar 23  for r >> R and rrr r  44rr 

 Now BrAr  in spherical coordinates for the magnetic dipole (with magnetic dipole  moment mIaIRz 2 ˆ ) is:

 o 2m Brr  3 cos 4 r

 o m  Br  3 sin valid for r >> R and rrr r  4 r  Br  0     BrBrrBrBrr ˆ   ˆ

   m   Bro 2cos rˆ sinˆ  mIaIRz 2 ˆ Thus:  3  for r >> R and rrr r  and . 4 r    cf w/ the E -field associated with a physical electric dipole with dipole moment p  qd :   1 p    Er2cos rˆ sinˆ  p  qd  3  for r >> d and rrr r  and . 4 o r

14 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

  We already know what this Br looks like it – it is “solenoidal” around the current loop:

Cross-Sectional View of a Magnetic Dipole Loop:

   2   mr  mIaIRz ˆ o  Thus if = a constant vector, Ar 3 for r >> R and rrr r  4 r    mr  o  Br Ar    3 for r >> R and rrr r  4 r     rr r Bro  m m    0  3  3  3 4 r r r    0

     rrmxˆ   rm33mr r mrˆˆ r m e.g. mmxx 3  m  xx33  533  5  3 x rr r rr r r  2  rrrr33333  3 r 33 and: 33 r  53   5   3  5   3  3 0 rr rr r r r r r

   Magnetic Field oomr 3mr ˆˆ r m  Br Ar   33    of a Magnetic 44rr Dipole Loop  for r >> R (far away) and rrr r 

 The magnetic field of a distant circuit (r >> R) does not depend on its detailed geometry,  but only its magnetic dipole moment, m !!!  Important (conceptual) result!!!

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 15 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

  Compare this result for Br for the magnetic dipole loop, with magnetic dipole moment    mIa , with result for the electric dipole field Er associated with a physical electric dipole   moment p  qd :

    o 3mrˆˆ r m    Br  3 mIa , for rR (far away) and rrr r  4 r    1 3prrˆˆ p      Er  3 p  qd , for rd (far away) and rrr r  4 o r +q When r ≈ R (or less) for magnetic dipole loop Or r ≈ d (or less) for electric dipole, then   will be able to “see” / observe / detect higher- d p  qd order moments - e.g. , octupole,  sextupole, etc. . .moments of the BE fields. −q

The statement that the magnetic field of a distant circuit (r >> R) does not depend on its detailed  geometry, but only its magnetic dipole moment, mIa (n.b. this is also true for the electrostatic   case, with p  qd ) are very useful!!!

    If one can compute mIa then one can obtain Ar and hence BrAr       (or if have p  qd then can obtain Er) for r >> R (or d) and rrr r  . EASY!!!

Magnetic Flux Conservation

      If BrAr  then Br Ar  0 is automatically satisfied everywhere r   If Br 0 for each/every point, r in a volume v bounded by its surface S   Then: Br d B  r ndAˆ (by the ) vS     What is Br ndAˆ ?? B r ndAˆ  0 S S   Recall Gauss’ Law for E (and/or D ) were:  D ndAˆ  Qenclosed = net electric displacement flux through closed surface S . DfreeS   E r ndAˆ  Qenclosed  = net through closed surface S . ETotoS

  Thus: B r ndAˆ 0 = net magnetic flux through closed surface S  0 !!! m S  Magnetic flux is conserved → magnetic field lines have no beginning / no end points (because  no magnetic charge(s)!) 2 The SI units of magnetic flux m are Tesla-m = Webers

16 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  Again: Do not confuse magnetic flux, m with the magnetic scalar potential m r (they even have the same units!!! Webers / Tesla-m2) WAA-HEE !!!

They are not the same thing!!!   Ar Ar m  r        Br ndAˆ BrArArr      m S m

Magnetic Flux Magnetic Vector Potential Magnetic Scalar Potential Area element

Don’t confuse these either!!!

       1 Ar    Ar 2   r   r rd gives Ar 0 mm m 4 v r

The magnetic flux through a surface S (not necessarily closed!!!):  Br ndSˆˆ   A  r ndS  A r d by Stoke’s Theorem m SS  C

n.b. not a closed surface!

 Magnetic flux enclosed by contour C :  Ard m C

 n.b. This d  is NOT a line segment associated with a line current I !!!

  B r ndSˆˆ   A  r  ndS  A r  d    r  d  m SS  CC mm

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 17 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Griffiths Example 5.11:

A spherical shell of radius R carries a uniform density  and rotates with   constant angular  , Determine the magnetic vector potential it produces at point r .

  A rotating surface charge density  produces a surface/sheet Kr  vr       Kr The magnetic vector potential is thus: Ar o da 4 S r

 For ease of integration, choose the observation/field point Pr Pzz ˆ  (i.e. rzz ˆ ) along the  zˆ -axis and  to lie in the x-z plane. Choose the origin  to be at the center of the sphere, as shown in the figure below:

 r rr  R22  r 2cos Rr  from the law of cosines

    Kr  vr  and: vr r   sinxˆ  cos zˆ (here)  rRsin cos xRˆˆ sin sin yR cos zˆ   vr r xyzˆˆˆ  vr   sin 0 cos RRRsin cos sin  sin cos 

 vr R cos  sin sin  xˆˆ   cos  sin cos  sin  cos  y   sin  sin sin  zˆ

22 Now since sindd cos   0 00   Then terms involving only sin or cos in the integral for Ar contribute nothing.

18 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

    Kr r A roo da da 44SSrr   with Kr  r  and da R2 sin d d and r Rr222cos Rr 

3   R sin  cos sin Then: Ar o d yˆ 0 22 2 Rr2cos Rr 

Let: u  cos  0 u  1 du sin d   u 1

 cos sin 1 udu d  0122 22 R rRr2cos RrRru  2 u1 RrRru22 Rr22  2 Rru 3Rr22 u1 1 22 22 22 R rRrRrRrRrRr       3Rr   2r If rR (i.e. inside sphere) then this integral =2 3R  2R If rR (i.e. outside sphere) then this integral = 2 3r   Now:  rrysin ˆ

  R   Then: Arro  for rR (inside sphere) 3  R4   Aro  r for rR (outside sphere) 3r3

  If we now rotate the problem so that   zˆ and rr ,,     then  rrysin ˆ  r sinˆ , thus with  rotated to   zˆ and field point now located    at rr ,, , the magnetic vector potential Ar inside/outside the rotating sphere becomes:     o R  ArR Ar,,  r sin ˆ ( rR , inside sphere) Ar max 3  1  R2sin  ~ r 3 o  R4 sin  2 o ˆ ~1 r Ar,,  2 ( rR , outside sphere) 3 r r rR

 zˆ  2R 22  Then: BrAr  o cos rˆ  sin   RzR ˆ    !!! (  zˆ ) 333oo

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 19 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Griffiths Example 5.12:

  Determine the magnetic vector potential Ar of an infinitely long solenoid with n turns / unit length, radius R and steady current I   I  n.b. The current extends to infinity, so we cannot use Ar o d because it diverges! 4 C r But we do know that:

  Magnetic flux Ar d   Ar   da  Bda  m CS  S

Flux-enclosing loop / contour

   Brda m S

But we know from Ampere’s Circuital Law that:  BinsiderR o nIzˆ = uniform & constant

inside 22  mo nI  R  o nI R

  But: inside Ar  R d where dRd  ˆ m C inside  m Ar  R2 R   Now Asolenoid must be parallel to I  Iˆ for the “ideal solenoid” (i.e. no pitch angle)   Ar Ar ˆ

 inside nI R 2 1 Then: Ar R mo  ˆˆ nIR 2 R 2 R 2 o

   2 A rR  1 R Ar max ArRnI ˆ 1 If r > R, then more generally, we have: outside o   2 onIR 2 r ~ r  ~1 r  1 For r < R, then: ArRnIrinside  o ˆ r 2 rR

  Note that: Ar A  rˆ (only) for the infinitely long ideal solenoid.

20 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

 Does BrAr  ?  1   BrArrArz    ˆ in cylindrical coordinates rr 0  2 11 R 11 2 BrRnIroutside o  znIRˆ  o zˆ  0 2 rr r 2 rr  11 2 11  Binside r R o nI r zˆˆˆ  o nI  2 r z  o nIz 22rr  r

     Does Ar 0 ?? (Coulomb Gauge) Ar A  rˆ

In Cylindrical Coordinates:     1 Ar   Ar  0 because Ar has NO explicit  -dependence! r 

 1 R2 ArRnIoutside o ˆ 2 r  1 ArR nIrˆ inside2 o

Magnetostatic Boundary Conditions

In the case of electrostatics, we learned (via use of Gauss’ Law -     E r ndaˆ  Qenclosed  ) that the normal component of Er suffers a discontinuity ETotoS whenever there is a surface charge density (free or bound) present on a surface / interface:

 above below  above below Tot free bound VV21 EE21   oonnsurface surface above above 21 (  is continuous across interface) n.b.  = perpendicular component relative to surface,  = parallel component relative to surface:

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 21 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Consider a thin conducting sheet of material carrying a surface current density of  KKxKxˆˆ   Amperes/meter

 Now imagine that this current sheet KKxKxˆˆ   is “placed” in an external magnetic field, e.g. created / emanating from some other current-carrying circuit below this current sheet.

 Call this external magnetic field that is below the original current sheet Bbelow . 1ext

  What we discover is that the magnetic field above the current sheet Babove is not parallel to Bbelow 2ext 1ext - it has been refracted by the current sheet (in the tangential direction - with respect to the surface)!

The physical origin for this is simple to understand. Below the current sheet, the current sheet  1  itself adds to the tangential component of Bbelow a component Bbelow   Kyˆ (for KKx ˆ ), ext sheet2 o above however, above the current sheet, the current sheet adds to the tangential component of Bext a  1  component Babove   Kyˆ (for KKx ˆ ). sheet2 o

 So if: B BxByBzˆˆ ˆ ext extxyz ext ext

 belowbelow below  below Then: BextBxByBz extˆˆ ext ext ˆ ║ = parallel to surface  xyz And: BaboveBxByBzaboveˆˆ above  above ˆ  = perpendicular to surface ext extxyz ext ext

22 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  Then by the principle of linear superposition, BBBTot ext sheet .  Hence, below the current sheet ( KKx ˆ ):

 below Bsheet   belowbelow below 1  below below below below BTOTBxB extˆˆˆˆ ext o KyBzBxByBz  extˆˆ  TOT TOT TOT xy2 z xyz    And above the current sheet ( KKx ˆ ):

 above Bsheet   aboveabove above 1  above above above above BTOTBxB extˆˆˆˆ ext o KyBzBxByBz  extˆˆ  TOT TOT TOT xy2 z xyz  

 above below Thus, (comparing BBTOT vs. TOT component-by-component), we see that:

below above  1) BBTOT TOT Tangential (to sheet / surface) component of BTOT parallel xx  BBbelow above to sheet current KKx ˆ is continuous. extxx ext

below above  2) BBTOT TOT Tangential (to sheet/surface) component of BTOT perpendicular y y  B aboveBK below  to sheet current KKx ˆ is discontinuous by an amount TOTyy TOT o

o K across sheet / surface.

 3) BBbelow above Normal (to sheet/surface) component of B is continuous TOTzz TOT TOT BBbelow above across sheet / surface. extzz ext

Mathematically, these 3 statements can be compactly combined into a single expression:   above below BTOTBKn TOT o ˆ where the unit normal to the surface, nzˆ  ˆ (here, as drawn above).

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 23 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  As we found in electrostatics, that the scalar electric potential Vr was continuous across    any boundary VrVrabove below , likewise, the magnetic vector potential Ar is also     continuous across any boundary, i.e. ArArabove below  provided that: Ar 0 , which  above below  guarantees that ArA  r and also provided that: Ar   Br , which, in   integral form, i.e. Ar d Br   da guarantees that ArArabove below . CS m

      However, note that the normal derivative of Ar, since Ar  K  r then Ar also   “inherits” the discontinuity associated with B r : B aboveBK below  (see #2 on previous  TOTyy TOT o  page), and since BrAr , thus we have a discontinuity in the (normal) slope(s) of   Aron either side of the boundary/current sheet.

  We can understand the origin of this condition on the normal derivative(s) of Ar taken just  above/below an “interface” e.g. for the specific case of the current sheet KKx ˆ . From  B aboveBK below  we know that the discontinuity in the B -field is in the yˆ -direction, TOTyy TOT o   whereas since the magnetic vector potential associated with the current sheet Aris always    parallel to the current, and since KKx ˆ we know that the component of ArTOT  that we are    concerned with here is in the yˆ -direction. But from: BTOTA TOT , then: BATOT TOT y y   thus we need to worry only about the yˆ -component of the curl of ArTOT , which is:   A A TOTx TOTz BATOT TOT   y y  zx

Then, noting that the zˆ -direction is perpendicular (i.e. normal) to the plane of the current sheet: AAaboveAAabove below below BBabove below  KTOTxx TOTzz  TOT  TOT TOTyy TOT o zx  zx surface surface above below above below above below AATOT TOT AA AATOT TOT xx TOTzz  TOT xx  zz  x  x nn surface  surface surface  0 A suffers no discontinuity TOTz Neither A nor A suffer discontinuities in their slopes at the current sheet – only TOTz TOTy A above does - in the normal (i.e. zˆ ) direction. Therefore, we can most generally write this TOTx   condition on the discontinuity in the normal derivative on Aras:  Arabove Ar below     K nno surface surface

24 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

  The Magnetic Vector Potential ArAssociated with a Finite Circular Disk Sheet Current

  We wish to delve a deeper into the nature of the magnetic vector potential, Arand also   BrAr associated with current sheets. Consider a sheet current KKx o ˆ flowing on the surface of a finite circular disk of radius R, lying in the x-y plane as shown in the figure below:

zˆ

z r

 yˆ

 xˆˆ, KKx o z

  To keep it simple, we’ll just calculate Arat an arbitrary point along the zˆ -axis above and   below the x-y plane. The magnetic vector potential Arassociated with a sheet current is:

    Kr Kxˆ da Arooo da 44SSrr

We deliberately chose a sheet current flowing on a finite circular disk of radius R so that we could easily carry out the integration. The area element da on the circular disk (in cylindrical coordinates) is da  d  d  d d , and from the figure above, we see that: r  22z .

 R  Kxˆ R 2  dd 2  Kxˆ  R  d 1 Azoo oo Kxˆ 22 z 4 00 22 4  0 222 oo Thus:   z   z  0 11 Kxˆˆ R22 z z 2 K R 22 z z 2 x 22oo oo

Now there is a subtlety here that we need to notice before proceeding further – since we are  interested in knowing Azat an arbitrary point along the zˆ -axis - above and/or below the x-y plane, thus z can be either positive or negative. Note that both the R22 z and z2 terms are always  0 for both positive and/or negative z (in particular: zzz2 !). Thus, in order to  preserve this fact, we explicitly keep expression for the magnetic vector potential Az as:

 1 Az K R22 z z 2 xˆ 2 oo 

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 25 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede  A plot of the magnetic vector potential Azvs. z is shown in the figure below for a circular  disk of radius R = 10 m and sheet current KKxo ˆˆ1.0 x Amperes/meter.

 Note that Azis a maximum when z = 0, right on the sheet current. Note also the discontinuity  in the slope(s) of Az on either side of z = 0, which arises due to the presence of the sheet current in the x-y plane, since:  Arabove Ar below     K nno surface surface  Azabove00  Az below   or:   K zzo zz00

Care/thought must also be taken when carrying out the normal derivatives (slopes) above and below the x-y plane – look carefully at the slopes for z > 0 and z < 0 in the above figure, and compare this information to what we calculate:

 above Az 0  zz   z  111K Rz22 zx 2ˆˆˆ K x K1 x 222oo  oo22 2 oo 22 zz Rz z Rz  below Az 0  zz   z  111K Rz22 zx 2ˆˆˆ K x K1 x 222oo  oo22 2 oo 22 zz Rz z Rz

26 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

Thus we see that indeed:  Azabove00  Az below    =11 Kxˆˆ Kx Kx ˆ K zz22oo oo oo o zz00  The magnetic field Bz at an arbitrary point along the along the zˆ -axis – either above and/or  below the x-y plane is calculated using BzAz in Cartesian coordinates. Since    Azx  Az A  zxˆ (only), then: BzAzAzxy  x ˆˆ  x z Thus:   above above above above Azx 0 1 z  B zAzAzxyK000  xoo  ˆˆ  2    1y ˆ z Rz22   below below below below Azx 0 1 z  B zAzAzxyK000  xoo  ˆˆ  2    1y ˆ z Rz22

 The figure below shows the magnetic field Bz vs. z along the zˆ -axis with a sheet current  KKx o ˆ flowing on the surface of the finite disk of radius R, lying in the x-y plane:

© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 27 2005-2008. All Rights Reserved. UIUC Physics 435 EM Fields & Sources I Fall Semester, 2007 Lecture Notes 16 Prof. Steven Errede

We now investigate what happens in the limit that the radius of the sheet current-carrying circular disc, R , i.e. it becomes an infinite planar sheet current. We discover that the    magnetic vector potential Arassociated with the sheet current KKx o ˆ becomes infinite   (i.e. Ardiverges):

 11 lim A z K R22 z zx 2ˆˆ K 22 z zx 2  oo  oo  R 22

  whereas the boundary condition on the discontinuity in the normal derivative of Aracross the sheet current lying in the x-y plane at z = 0 still exists, and is well-behaved (i.e. finite):

 Azabove00  Az below    =11 Kxˆˆ Kx Kx ˆ K zz22oo oo oo o zz00

  We also discover that the magnetic field Br is also well-behaved (i.e. finite) – and constant – independent of the height/depth z above/below the x-y plane (!!):

 z limBabove zK 011  1 yKyˆˆ  22oo22 oo R z  z limBbelow zK 011  1 yKyˆˆ  22oo22 oo R z

28 © Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois 2005-2008. All Rights Reserved.