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.
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The basic force of magnetostatics
1. Recall electrostatics.
F12 =
q1q 2 r1 r2
3
4QF0 r1 r2
2. In magnetostatics there is an equivalent empirical law which we take as a
postulate
F12 =
q1q 2N0 v1 × ¢ v2 × ( r1 r2 )¯±
3
4Q
r1 r2
3. The force is proportional to 1/r 2 .
4. The force is perpendicular to v1 : F12 ¸ v1 = 0 .
5. The force depends on the particle velocities v1, v2
6. The constant of proportionality is equal to N0 = 4Q × 107 Henry ' s / meter .
The magnetic field
1. We define the magnetic field in terms of the force as follows
2
Electrostatics
F12 =
Magnetostatics
F12 =
q1q 2 r1 r2
3 w q 1E
4QF0 r1 r1
r1 r2
q
E= 2
3
4QF0 r1 r1
N0q1q 2 v1 × ¢ v 2 × ( r1 r2 )¯±
w q 1v1 × B
3
4Q
r1 r2
B=
N0q 2 v 2 × ( r1 r2 )
3
4Q
r1 r2
2. Let’s extend the definition to include many charges
Electrostatics
E ( r) = œ
q j r rj
3
4QF0 r rj
B ( r) = œ
N0q j v j × ( r rj )
3
4Q
r rj
j
Magnetostatics
j
3. Let’s extend the analysis to a continuum
Electrostatics
œq
¨ S d ra
l
j
j
E ( r) =
Magnetostatics
œq v
j
1
4QF0
l
j
j
r ra
¨ S r ra
3
d ra
¨ S vd ra = ¨ J d ra
J = S v = current density in Amps/m2
B ( r) =
N0
4Q
¨
J ( ra ) × ( r ra )
r ra
3
d ra
4. Now, recall that
r ra
1
1
= +‹a
3 = ‹
r ra
r ra
r ra
5. This allows us to introduce the vector potential A
3
Electrostatics
E ( r) =
1
4QF0
=
¨
1
4QF0
= ‹
r ra
a
3 dr
r ra
S
1
¨ S‹ r ra d ra
1
4QF0
1
¨ S r r a d ra
= ‹G
G=
Magnetostatics
B ( r) =
1
4QF0
N0
4Q
=
=
1
¨ S r ra d ra
¨
N0
4Q
N0
4Q
J ( r a) × ( r ra)
¨
= ‹×
r ra
3
d ra
1
¨ J (ra) ׋ r ra d ra
‹×
J ( r a)
d ra
r ra
J ( r a)
N0
d ra
4 Q ¨ r ra
= ‹×A
A=
N0
4Q
¨
J ( ra)
d ra
r ra
6. A is the vector potential and the integral expression for A is known as the BiotSavart law.
7. A basic property of magnetostatics
Electrostatics
E = ‹G
l
‹×E = 0
Magnetostatics
B = ‹×A
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
s
sS
(S%x %y %z ) = %x %y %z
st
st
(2) = flux of charge × area = Sv %y %z x %x / 2
(1) =
(3) = flux of charge × area = Sv %y %z x +%x / 2
6. Recall that S v = J = current density . The conservation law becomes
%x %y %z
(
= %y %z (J
)
)
sS
= %y %z Sv x %x / 2 Sv x +%x / 2
st
x x %x / 2
Jx
x +%x / 2

sJ % x
sJ %x ¬­
x %y %z žžJ x (x ) x
J x (x ) x
­
Ÿ
sx 2
sx 2 ®­
sJ x
= %x %y %z
sx
7. Therefore, in 1-D
sS sJ x
+
=0
st
sx
8. In 3-D
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sJ
sS sJ x
sJ
+
+ y + z =0
st
sx
sy
sz
sS
+‹¸J = 0
st
9. Note, in electrostatics conservation of charge implies sS / st = 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
A=
N0
4Q
¨
J ( r a)
d ra
r ra
‹¸A =
N0
4Q
¨
‹¸
J ( ra )
d ra
r ra
¯
a
¡‹a ¸ J ( r ) + 1 ‹a ¸ J ( ra)° d ra
¨¡
°
r ra
r ra
¡¢
°±
J ( ra )
N
d ra
= 0 ¨ ‹a ¸
4Q
r ra
=
N0
4Q
=
N0
4Q
¨
n a ¸ J (S a)
dS a
r ra
= 0 for a localized current distribution: J (S a) = 0 as S a l d
‹¸A = 0
The analog of Poisson’s equation
1. In electrostatics we have
E = ‹G
‹2G = S / F0
2. In magnetostatics
6
A=
N0
4Q
¨
J ( ra )
d ra
r ra
1
N0
J ( r a ) ‹2
d ra
¨
4Q
r ra
N
= 0 ¨ J ( ra) 4QE ( r ra)d ra
4Q
2
‹ A = N0 J
‹2 A =
3. This is the equivalent to Poisson’s equation.
4. Also, we find that
‹ × B = ‹ × (‹ × A)
= ‹ (‹ ¸ A ) ‹2 A
= ‹2A
‹ × B = N0 J
5. This is Ampere’s law for magnetostatics.
6. We can now easily obtain the integral formulation of magnetostatics. First
equation.
¨ ‹ ¸ Bd r = 0
l
V
¨ n ¸ B dS = 0
S
7. Second equation.
¨ (‹ × B N J) ¸ n dS = 0
0
S
l
¨v B ¸ d l =N ¨ J ¸ n dS
0
C
S
7
Summary and comparison
Magnetostatics
N0
4Q
Integral relation
A=
Differential formulation
B = ‹×A
¨
J ( ra)
d ra
r ra
‹2A = N0 J
Integral formulation
Force density
Electrostatics
G=
1
4QF0
¨
S ( ra )
d ra
r ra
E = ‹G
‹2G = S / F0
¨ n ¸ B dS = 0
¨v E ¸ d l = 0
¨v B ¸ d l = ¨ J ¸ n dS ¨ n ¸ E dS = ¨ S d r
F=
qE
%r
= SE
F=
qv × B
%r
= J×B
Example 1
1. Find the magnetic field in a long solenoid.
2. By symmetry there is only a z component of field: B = Bz ez
3. With no sources at infinity Bz must vanish outside the coil: Bz = 0
for r > a .
4. By symmetry Bz (r , R, z ) = Bz (r )
5. Inside the solenoid
‹ × B = N0 J
sBz
= N0J R = 0
sr
6. The solution is
Bz = B0 = const .
7. Express B0 in terms the current flowing in the wire I by using Amperes law.
8
¨v B ¸ d l = N ¨ J ¸dS
0
B0L = N0NI
8. Therefore, inside a long solenoid the field is uniform with a value
B0 =
N0NI
L
Example 2
1. Find the field inside a torus.
2. By symmetry we see that s / sG = 0
3. The only non-zero field component is B = BG eG .
4. As in the straight solenoid BG = 0 outside the torus.
5. Inside the torus we have
‹ × B = N0 J
1 sRBG
=0
R sR
6. The solution is
BG =
K
R
9
7. Express K in terms of the coil current by using the integral form of Ampere’s
law
¨v B ¸ d l = N ¨ J ¸ n dS
0
2QRBG = N0NI
8. This gives
BG (R ) =
N0NI
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
A=
N0
4Q
¨
J ( ra)
d ra
r ra
4. Note that the current density is given by
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J ( ra) = I E (R a a ) E (Z a) eG
¨ J ¸ n dS a = ¨ I E (Ra a ) E (Z a)dRadZ a = I
5. One tricky problem is that eGa v eG . 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,
eG = sin G ex + cos G ey
eGa = sin Gaex + cos Gaey
7. The Biot-Savart law becomes
A = Ax ex + Ay ey
=
N0
4Q
=
N0
4Q
Ax =
N0I
4Q
Ay =
N0I
4Q
¨
J Ga eGa
d ra
r ra
¨
J Ga ( sin Gaex + cos Gaey )
d ra
r ra
8. We see that
¨
E (R a a ) E (Z a)( sin Ga)
d ra
r ra
¨
E (R a a ) E (Z a)(cos Ga)
d ra
r ra
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9. The next step is to combine these terms to evaluate AG ( r) = Ax sin G + Ay cos G
at the observation point.
AG =
N0I
4Q
=
N0I
4Q
¨
E (R a a ) E (Z a)
(cos G cos Ga + sin G sin Ga)d ra
r ra
¨
E (R a a ) E (Z a) cos (Ga G)
d ra
r ra
10. To evaluate this integral note that
1/ 2
2
2
2
r ra = ¡ (x a x ) + (y a y ) + (z a z ) °¯
¢
±
2
2
2 1/ 2
= ¡(R a cos Ga R cos G ) + (R a sin Ga R sin G ) + (Z a Z ) ¯°
¢
±
1/ 2
2
= ¡R 2 + R a2 2RR a cos (Ga G ) + (Z a Z ) ¯°
¢
±
11. We can now carry out the integration over Ra, Z a .
AG =
=
N0I
4Q
¨
N0Ia
4Q
¨
E (R a a ) E (Z a) cos (Ga G)
R adR adZ ad Ga
r ra
cos (Ga G)
R 2 + a 2 2Ra cos (Ga G) + Z 2 ¯
¡¢
°±
1/ 2
d Ga
12. This integral can be evaluated in terms of the complete elliptic integrals
K (k ) , E (k ) . 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 R = r sin R
Z = r cos R and assume r a .
12
15. In this limit
1
R 2 + a 2 2Ra cos (Ga G) + Z 2 ¯
¡¢
°±
1/ 2
1
=
r 2 + a 2 2ra sin R cos (Ga G)¯
°±
¡¢
1 a
¯
x ¡1 sin R cos (Ga G)°
°±
r ¡¢
r
1/ 2
16. The vector potential becomes
AG =
=
a
N0Ia cos (Ga G ) ¯
¡1 sin R cos (Ga G )° d Ga
¨
¡¢
°±
r
r
4Q
2
N0 (Qa I ) sin R
4Q
r2
17. Comparison with electrostatics:
qd sin R
P sin R
=
2
4QF0 r
4QF0 r 2
P = qd = dipole moment
Electric dipole:
G=
Magnetic dipole:
AG =
N0 (Qa 2I ) sin R
N0M sin R
4Q
r
4Q r 2
M = Qa 2I = AI = magnetic moment
2
=
18. A circular loop of wire produces a magnetic dipole field
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