Lecture 6 Basics of Magnetostatics

Today’s topics

1. New topic – magnetostatics 2. Present a parallel development as we did for electrostatics 3. Derive basic relations from a fundamental force law 4. Derive the analogs of E and G 5. Derive the analog of Gauss’ theorem 6. Derive the integral formulation of magnetostatics 7. Derive the differential form of magnetostatics 8. A few simple problems

General comments

1. What is magnetostatics? The study of DC magnetic fields that arise from the motion of charged particles. 2. Since charged particles are present, does this mean we also must have an electric field? No!! Electrons can flow through ions in such a way that current flows but there is still no net charge, and hence no electric field. This is the situation in magnetostatics

3. Magnetostatics is more complicated than electrostatics. 4. Single magnetic charges do not exist in nature. The simplest basic element in magnetostatics is the magnetic dipole, which is equivalent to two, closely paired equal and opposite charges in electrostatics. In magnetostatics the dipole forms from a small circular loop of current. 5. Magnetic geometries are also more complicated than in electrostatics. The vector nature of the magnetic field is less intuitive.

1 The basic force of magnetostatics

1. Recall electrostatics.

qq rr = 12 1 2 F12 3 4QF0 rr12

2. In magnetostatics there is an equivalent empirical law which we take as a postulate

qq vv×× ¯( rr ) = 12N 0 1212¢± F12 3 4Q rr12

3. The force is proportional to 1/r 2 . = 4. The force is perpendicular to v1 : Fv12¸ 1 0 .

5. The force depends on the particle velocities vv12, =×7 Henry s meter 6. The constant of proportionality is equal to NQ0 410 '/ .

The magnetic field

1. We define the magnetic field in terms of the force as follows

2 qq rr = 12 1 2 q Electrostatics FE123 w 1 4QF0 rr11 q rr = 212 E 3 4QF0 rr11 qq vv×× ¯() rr = N012 1212¢±q × Magnetostatics Fv123 w 1 1 B 4Q rr12 q vrr×() = N02 212 B 3 4Q r1  r2

2. Let’s extend the definition to include many charges

q jjrr Electrostatics Er()= œ 3 j 4QF0 rr j × N q j vrrjj() Magnetostatics Br()= 0 œ 3 j 4Q rr j

3. Let’s extend the analysis to a continuum

qd Electrostatics œ j l S ra j ¨ 1 rr a Er()= d ra ¨ S 3 4QF0 rr a qd= d Magnetostatics œ jjvvrJ l S aar j ¨¨ Jv==S current density in Amps/m2 Jr()aa× ( r r ) ()= N0 d Br3 ra 4Q ¨ rr a

4. Now, recall that

rr a = 11=+ 3 ‹ ‹a rr a rraa rr

5. This allows us to introduce the vector potential A

3 1 rr a Electrostatics Er()= d ra ¨ S 3 4QF0 rr a 11 = ‹¨ S dra 4QF0 rr a 11 = ‹ ¨ S dra 4QF0 rr a = ‹G 11 GS= ¨ dra 4QF0 rr a Jr()aa× ( r r ) ()= N0 d Magnetostatics Br3 ra 4Q ¨ rr a N 1 =  0 J()rraa׋ d 4Q ¨ rr a N Jr()a = 0 ‹× dra 4Q ¨ rr a N Jr()a = ‹× 0 dra 4Q ¨ rr a = ‹×A

N Jr()a Ar= 0 d a 4Q ¨ rr a

6. A is the vector potential and the integral expression for A is known as the Biot- Savart law. 7. A basic property of magnetostatics

Electrostatics EE= ‹G l ‹×= 0 Magnetostatics BA= ‹× l‹¸ B= 0

A new relation – conservation of charge

1. A basic physical property is that charge is conserved. It cannot be created or destroyed. 2. This implies a relation between S and J . 3. Let’s derive this relation for a simple 1-D case

4 4. Statement of charge conservation

Rate of increase of charge = flow in of charge  flow out of charge (1) = (2)  (3)

5. The separate terms are as follows

ssS (1) = ()S%%%xyz= %%% xyz sstt =×vyz (2) flux of charge area = S %% xx% /2 =×vyz (3) flux of charge area = S %% xx+% /2

6. Recall that SvJ= = current density . The conservation law becomes

xyzsS = yz v v %%% %%()SSxx  xx+ st % /2% /2 = yzJ() J %%xxxx% /2  xx+% /2 ¬Jx Jx yzJx() s%xx Jx() s% x% %ž xx    ­ Ÿ®ž ssxx22­

sJx = %xyz % % sx

7. Therefore, in 1-D

ssS J x +=0 sstx

8. In 3-D

5 J ssS JJxzs y s +++=0 sstx s y s z sS + ‹¸J = 0 st

9. Note, in electrostatics conservation of charge implies ssS / t = 0 . This is a trivial result since we are looking at static situations with no time variation 10. However, in magnetostatics conservation of charge implies that ‹¸J = 0. The current is divergence free for static problems. 11. This implies something about ‹¸A as follows

N J(ra) Ar= 0 d a 4Q ¨ rr a N Jr()a ‹¸Ar= 0 ‹¸ d a 4Q ¨ rr a

N ¯Jr()a 1 = 0 ¡°‹aaaa ¸ + ‹¸Jr()d r 4 ¨ ¡°rraa rr Q ¢±¡° N Jr()a = ‹¸0 aadr 4Q ¨ rr a N nJaa¸ ()S =  0 dSa 4Q ¨ rr a ==0 for a localized current distribution: J()SSaa 0 as ld ‹¸A = 0

The analog of Poisson’s equation

1. In electrostatics we have E = ‹G 2 = ‹ GS / F0

2. In magnetostatics

6 N J()ra Ar= 0 d a 4Q ¨ rr a N 1 ‹22AJrr= 0 ()aa‹ d 4Q ¨ rr a N = 0 J()rrraa4QE ( )dra 4Q ¨ 2 = ‹ AJN0

3. This is the equivalent to Poisson’s equation. 4. Also, we find that

‹×=BA‹×()‹× = ‹‹¸()AA ‹2 = ‹2A ×= ‹ BJN0

5. This is Ampere’s law for magnetostatics. 6. We can now easily obtain the integral formulation of magnetostatics. First equation.

¨¨‹¸Brdd= 0 l¸ nBS = 0 VS

7. Second equation.

( × BJn) dS= 0 Bl d= Jn dS ¨¨‹ ¸NN00l¸v ¨ ¸ SCS

7 Summary and comparison

Magnetostatics Electrostatics

N Jr()aa1 S() r Integral relation Ar==0 ¨¨ddaaG r 44QQFrraa0 rr Differential formulation BA= ‹× E = ‹G 22= = ‹ AJ‹NGSF00  / Integral formulation ¨¨nB¸ dS= 0 v E¸ d l= 0 ¨¨v Bl¸ddSdSd= Jn¸¸ ¨ nE= ¨S r qqEvB× Force density FEF==S = =× JB %%rr

Example 1

1. Find the magnetic field in a long solenoid.

2. By symmetry there is only a z component of field: Be= Bzz

3. With no sources at infinity Bz must vanish outside the coil: Brz =>0 for a .

() 4. By symmetry Brzz(),,R z= Br 5. Inside the solenoid

×= ‹ BJN0

sBz = N J = 0 sr 0 R

6. The solution is

B== B const z 0 .

B I 7. Express0 in terms the current flowing in the wire by using Amperes law.

8 Bldd= JS ¨¨v ¸ N0 ¸ BL= NI 00N

8. Therefore, inside a long solenoid the field is uniform with a value

NI B = N0 0 L

Example 2

1. Find the field inside a torus.

2. By symmetry we see that ss/0G = = B 3. The only non-zero field component is BeGG . B = 4. As in the straight solenoid G 0 outside the torus. 5. Inside the torus we have

×= ‹ BJN0 1 sRB G = 0 RRs

6. The solution is

K B = G R

9 7. Express K in terms of the coil current by using the integral form of Ampere’s law

Bldd= JnS ¨¨v ¸ N0 ¸ RB= NI 2QNG 0

8. This gives

NI BR()= N0 G 2QR

Example 3

1. Find the field due to the current flowing in a loop of wire

2. This problem is more complicated than you would think. Proceed as follows. 3. Since there are no conductors and the current is localized it makes sense to start with the Biot-Savart law

N J()ra Ar= 0 d a 4Q ¨ rr a

4. Note that the current density is given by

10 ( ) = IR( a) ( Z) J reaaaEE G ¨¨Jn¸ dSaaaaa= IEE()() R a Z dR dZ= I

5. One tricky problem is that eeGa v G . As we integrate around the loop the

direction of eGa is constantly changing while at the fixed observation point eG does not change. 6. This difficulty is avoided by expressing the unit vectors in rectangular coordinates which always stay fixed. Thus,

= + eeG sinGGxy cos e = + eeGa sinGGaaxy cos e

7. The Biot-Savart law becomes

Ae=+AAxx yy e J N aae = 0 GGdra 4Q ¨ rr a J ()+ N a sinGGaaeexy cos = 0 G dra 4Q ¨ rr a

8. We see that

I ()()()Raaaa Z sin N0 EE G Adx = ra 4Q ¨ rr a I ()()()Raaaa Zcos N0 EE G Ady = ra 4Q ¨ rr a

11 AAA()= + 9. The next step is to combine these terms to evaluate G r  xysinGG cos at the observation point.

N I EE(Raaa ) ( Z) Ad=+0 ()cosGG cosaa sin GG sin ra G 4Q ¨ rr a N I EE()()()Raaaa Zcos GG = 0 dra 4Q ¨ rr a

10. To evaluate this integral note that

2221/2 rr aa= ¯()()()xx + yy a + zz a ¢±¡° 222 1/2 = ¯()()(RRaacosGG cos+ RR aa sin GG sin + ZZ a ) ¢±¡° 2 1/2 =+ ¯RR22aaaa2cos RR()()GG+ ZZ ¢±¡°

11. We can now carry out the integration over RZaa, .

N I EE(Raaaa) ( Z)cos( GG) A= 0 Raaaa dR dZ dG G 4Q ¨ rr a Ia cos()GGa  = N0 d 1/2 Ga 4Q ¨ ¯Ra22+ 2cos Ra( a ) + Z 2 ¢±¡°GG

12. This integral can be evaluated in terms of the complete elliptic integrals Kk( ), Ek( ) . This does not provide too much insight. 13. A more useful approach is to evaluate the integrals far from the loop of wire. 14. As shown below we let Rr==sinR Zr cosR and assume ra .

12 15. In this limit

11= 1/2 1/2 ¯ Ra22+ 2cos Ra()aa++ Z 2 ra 22 2sincos ra ()¯ ¢±¢¡°¡GG R GG±° 1 ¯a x¡°1sincosRGG()a  rr¢±¡°

16. The vector potential becomes

Iacos(GGa  ) a Ad= N0 ¯() G ¡°1sincosRGGGaa 4Q ¨ rr¢±¡° NQ()aI2 = 0 sin R 4Q r 2

17. Comparison with electrostatics:

qdsinRR P sin Electric dipole: G == rr22 44QF00 QF Pqd== dipole moment ()aI2 NQ0 sinRNM sin R Magnetic dipole: A ==0 G 44QQrr22 MaIAI===Q 2 magnetic moment

18. A circular loop of wire produces a magnetic dipole field

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