Download Current and resistance

Physics 460
Fall 2006
Susan M. Lea
Letzs start o¤ by reviewing what we accomplished in Physics 360. We
learned how to investigate the behavior of electromagnetic systems that are
constant in time. We derived Maxwellzs equations in the static limit. They
are:
~ ¢E
~ = ½
Coulombzs law (Gausszs Law) :
r
(1)
"0
~
~ ¢B
~ =0
Gausszs Law for B
:
r
(2)
~ £E
~
r
=
Amperezs Law
:
0
~ £B
~ = ¹ ~j
r
0
(3)
(4)
~ and the equations
We used these equations to derive the potentials V and A
they satisfy:
~ £E
~
r
~
~
r¢E
~ ¢B
~
r
~ £B
~
r
=
=
~ = ¡rV
~
= 0)E
½
½
2
=
)r V =¡
"0
"0
~ =r
~ £A
~
0)B
³
´
~ r
~ ¢A
~ ¡ r2 A
~ = ¡¹0~j
¹0~j ) r
Here we add the Gauge condition
Coulomb Gauge:
to obtain
~ ¢A
~=0
r
~ = ¹ ~j
r2 A
0
~ as well as V satis¤es
If we use Cartesian components, each component of A
Poissonzs equation:
r2 (function) = (source)
In a region where the sources are zero, we have Laplacezs equation:
r2 (function) = 0
and we learned how to solve this equation making use of boundary conditions
at the edges of the region.
We also derived expressions for the electric energy density:
uE =
1
"0 E 2
2
and the charge conservation equation
@½ ~ ~
+r¢j= 0
@t
(5)
~ = "E
~ and H
~ = B=¹
~
Finally, we showed how to use the auxiliary ¤elds D
to
simplify discussion of ¤elds in media.
Now it is time to expand our discussion to allow for time variation in the
sources and the ¤elds: we are to study Electrodynamics.
1
1
Current and resistance
1.1
Current
We start with some practical examples of systems where charges are moving.
Every current consists of moving charges. As we found in Physics 360 (see
360notes12)
~j = ne~v
where n is the number density of charged particles, e the charge per particle, and
~v the velocity of each particle. Conductors are systems than contain charges
free to move under the application of applied forces. Such systems usually
exhibit resistance. Like mechanical systems with friction, a force is needed to
start a particle moving, and the moving particles may reach a constant speed
with the applied force balancing the frictional force. The electrical systems
reach a state with constant current
~
~j = ¾ F = ¾ f~
q
where
F~
f~ =
q
is the force per unit charge. The constant of proportionality ¾ is the conductivity
¾=
1
½
and ½ is the resistivity. The resistivity of a good conductor like copper is
around 10¡ 8 -¢m; for semiconductors like silicon the value is around 10 3 ; while
for insulators like wood the value is around 10 11 :
While any force can in principle cause a current, we are going to be interested
in electromagnetic forces:
³
´
~ + ~v £ B
~
F~ = q E
so that
³
´
~ + ~v £ B
~
~j = ¾ E
(6)
This relation is often called Ohmzs law, although strictly it is not. What
Ohm actually stated is that resistance is independent of current. This result
follows from equation (6) if ¾ is independent of current. This is almost true
for materials like copper, as we shall see. In most ordinary conductors we may
~ term because v is small (¿ c); and the e¤ects of B
~ are
often ignore the ~v £ B
~ (as in the Hall e¤ect- see below).
often balanced by additional components of E
2
1.2
Resistance
Letzs ¤nd the relation between the resistance of a circuit component and its
fundamental properties (shape, size, material). Let the resistor be made of
~ ) and a
uniform material of resistivity ½; and have a length ` (parallel to E
cross-sectional area A that is constant along its length. Then the current may
be expressed in terms of the current density as
I = jA
(Remember: current = ¢ux of current density)
I = jA = ¾ EA
Now we may express E (assumed uniform along the length `) in terms of the
potential di¤erence:
¢V
E=
`
and so
¾A
I=
¢V
`
or
µ ¶
`½
¢V = I
= IR
(7)
A
a more familiar version of "Ohmzs Law". Thus the resistance is
R=½
`
A
(8)
It is easy to understand this result: like water ¢owing down a pipe, current ¢ows
more easily when the area of the "pipe" (the circuit component) is greater, and
less easily when the length is greater.
We made quite a few assumptions here, and we should verify them. First,
is the electric ¤eld uniform? We would set up the circuit by applying a ¤xed
potential di¤erence ¢V using a battery or generator. Once the system reaches
equilibrium, which it will do in a time of order `=c; that leaves us with a static
boundary value problem to solve. Letzs ground one end of the resistor for
simplicity. Then we have
V
=
0 at z = 0
V
=
V 0 at z = `
and
r2 V = ¡
½q
"0
where I have used the subscript q on the charge density to distinguish it from
resistivity. But what is ½q ? Well, we found that in a static situation without
currents, the charge density is zero inside a conductor. If the currents are
3
steady (we have reached equilibrium) then the charge density, zero or not, is
also constant. But then, from charge conservation (eqn 5)
@½q
@t
³ ´
~ ¢ ~j = ¡r
~ ¢ ¾E
~
= 0 = ¡r
and if the material is uniform so that ¾ = constant, then, using equation (1)
³ ´
½
~ ¢ ¾E
~ = ¾r
~ ¢E
~ =¾ q
0=r
"0
Then the charge density is zero, and so V satis¤es Laplacezs equation. In
addition, current ¢ows along the cylinder, but not across the boundaries into
the space outside, so
~j ¢ ^n = ¾ E
~ ¢n
~ =0
^ = ¡¾ ^n ¢ rV
Thus either V or its normal derivative is known everywhere on the surface of
the resistor, so the uniqueness theorem tells us that there is one unique solution
for the potential. A simple solution that satis¤es all the conditions is
V = V0
~
with a uniform E
z
`
~ = ¡rV
~ = ¡V 0 z^
E
`
as we assumed above.
We have proved a very powerful theorem:
In a material of uniform conductivity carrying a steady current, the charge
density is zero.
Note that this does not prohibit non-zero surface charge density on the
boundaries, and in general these surface charges are necessary for a self-consistent
solution.
Now letzs investigate what happens when the cross-sectional area is not a
constant. Suppose the circuit component is a cylindrical shell with radii a < b;
and we apply the potential di¤erence so that the inner surface at s = a has
potential V 0 and the outer surface at s = b has potential zero. Then the
appropriate solution of Laplacezs equation is (360notes8 page 8)
V = C 1 ln s + C2
To get V = 0 at s = b and V = V0 at s = a; we choose the constants as follows:
V =
with electric ¤eld
V0
½
ln
ln a=b b
~ = ¡ rV
~ = ¡ V0 s^ = V0 s^
E
ln a=b s
ln b=a s
4
~ points
where the second expression has ln b=a > 0; and thus shows that E
outward. The total current through a cylindrical surface of radius s
Z
Z ` Z 2¼
~ ¢ s^ sdµdz
~
I =
j¢n
^ dA = ¾
E
0
0
Z
V0
sdµ
V0 `
= ¾
`
=¾
2¼
ln b=a
s
ln b=a
The result is independent of s; showing that the current is constant throughout
the resistor, as expected. The resistance is
R=
¢V
V0
b
ln b=a
=
ln = ½
I
2¼¾`V 0 a
2¼`
Gri¢thsz Example 7.2 introduces the charge per unit length on the cylinder,
which is unnecessary and involves the additional assumption that the electric
¤eld inside the inner cylinder is zero. If the ¤eld for s < a is zero, we could use
our solution to ¤nd ¸ from the given value of V0 :
Both Gri¢ths and LB discuss the "Drude" model for conductivity, and you
should de¤nitely look at it, bearing in mind that modern quantum theories are
quite a bit di¤erent. Graduate students should look up what Feynmann has to
say on the subject.
As a result of the many collisions undergone by the moving electrons, the
work done by the battery increases the thermal energy of the resistor. The
amount of charge passing through potential di¤erence ¢V in time t is
Q = It
and the work done by the battery is
W = Q¢V = It¢V
thus the power delivered to the charges is
P =
W
R2
= I¢V = I2 R =
t
¢V
This power serves to heat the resistor, thus increasing the temperature and
changing the resistance. This e¤ect is used to advantage in electric toasters,
hair dryers, and light bulbs, but is a nuisance in TV sets, for example. It is
sometmes called Joule heating.
We should remember several important facts from this discussion of steady
currents in circuits. First, each circuit has a self-consistent distribution of
charge, with resulting ¤elds and potential di¤erences. Charge densities occur
only where the electrical properties changew usually at the surface of wires,
the ends of resistors, and similar places. As we saw with conductors in Phys
360, each charge moves a tiny distance to establish the equilibrium, and the
timescale to establish equilibrium is the time for the ¤elds to adjust, equal to
(length scale of system)/(speed of light). The self-consistent ¤elds serve to
distribute the e¤ects of batteries, for example, around the circuit.
5
2
2.1
EMF
Batteries
The in¢uence of the external agent that drives the circuit (battery, photvoltaic
cell, whatever) is often described by a quantity called EMF or elctro-motive
force. It is not a force at all, but the line integral of the force per unit charge
around the circuit:
I
E = f~e xt ¢ d~`
Once the circuit reaches equilibrium (very fast! see above), the self-consistent
electrostatic ¤elds donzt contribute, because (eqn 3)
I
~ £E
~ = 0 ()
~ ¢ d~` = 0
r
E
Thus E re¢ects the contribution of the external agent.
For an ideal battery with zero internal resistance, the net force on the charges
inside the battery is zero. This force has two components, one electrical and
chemical, so they have to balance.
Z
te rm in al b
te rm in al a
³
´
~ + f~ex t ¢ d~` = 0
E
So the potential di¤erence across the battery is
Va ¡ V b
=
Z
t erm in al b
ter min al a
Z te rm in al b
= ¡
=
I
~ ¢ d~` = ¡
E
Z
te rm in al a,in sid e b att ery
te rm ina l b
te rm in al a
f~ex t ¢ d~`
f~e x t ¢ d~` ¡
Z
te rm in al b
te rm in al a,ou tsid e b atte ry t
f~e xt ¢ d~`
f~e xt ¢ d~` = E
where we used the fact that fe xt = 0 outside the battery to add the extra (zero)
term in line 2. See LB §26.1 for batteries with internal resistance.
2.2
Motional emf
Motional emf arises whenever a conductor moves through a magnetic ¤eld, and
is the basis for simple generators. To see how it works, letzs consider a simple
rectangular loop with a resistance R on one side, as in Gri¢ths Figure 7.10.
6
On side ab; every electron in the wire experiences a magnetic force per unit
charge
~ = vB [^x £ (¡^
f~ma g = ~v £ B
z )] = vB y^
Thus we have an EMF at the instant shown of
I
E = f~m ag ¢ d~` = vBh
where the integral was taken clockwise around the loop. There is no contribution
~ = 0 or ~v £ B
~ is perpendicular
from other segments of the loop because either B
~
to dl: This emf drives a current clockwise around the lo op. The segment ab
then experiences a magnetic force due to the current of
F~ =
Z
a
b
~ = IhB (¡^
Id~l £ B
x)
Segments bcand da experience forces too, but they are equal and opposite, and
sum to zero. Thus the net force on the loop is in the minus-x direction, and
if nothing else is done the motion of the loop stops, the EMF! 0; and the
current! 0: But if we pull the lo op to the right with a balancing force
F~ p u ll = ¡F~ m ag = IhB x^
we can keep the motion going. We have to do work at a rate
P = F~p u ll ¢ ~v = IhBv
The electrical power expended in the circuit is
P = I 2 R = IE = IvBh
7
The two powers are equal, as they must be. Remember: magnetic force does
no work!
Motional emf o ccurs whenever the size, shape or orientation of a loop changes
and a magnetic ¤eld is present. It is one example of Faradayzs law, which may
be expressed in the more general form
¯
¯
¯ d© B ¯
¯
¯
jEj = ¯
(9)
dt ¯
(Notice the absolute value signs!) The magnetic ¢ux through the loop is
Z
~ ¢n
©B =
B
^ dA
where the integral is over a surface spanning the loop and n
^ is the usual normal
to the surface element dA: In our case, with normal chosen into the paper (the
~ we have
¡^z direction, parallel to B)
©B = Bh (¡x)
(Notice that x is the location of side ab with respect to the edge of the magnetic
¤eld region, and is negative). Thus
d©B
dx
= ¡Bh
= ¡Bhv
dt
dt
Thus we get the right value for jEj : Putting the signs back, we have
I
Z
d© B
d
~ ¢ ^n dA
E = f~m ag ¢ d~` = ¡
=¡
B
dt
dt
(10)
where now the directions of n
^ and d~` are related through the usual right hand
rule.
~ changes, n
The time derivative of ¢ux is non-zero if B
^ changes or the area
~
changes. In our case the area with non-zero B is decreasing. But when B
changes we donzt call it motional emf any more.
Equation (10) is the integral form of Faradayzs law, and it includes the
~ as well as the motional emfzs we have
case of a static loop with changing B
already discussed. To use it correctly, you must choose a direction for d~` (that
is, a direction for going around the loop) and then choose the direction for the
normal n
^ according to the right hand rule. Equivalently, you can use equation
(9) to relate the magnitudes of the emf and the rate of change of ¢ux, and use
Lenzzs law to get the directions right.
Lenz0 s law:
The induced emf opposes the change that creates it.
Lenzzs law is a statement of energy conservation: without it we could build
perpetual motion machines that would run forever without any energy input.
8
When the loop is stationary but the magnetic ¤eld changes, there is no
magnetic force because ~v = 0: So what causes the emf? It is an electric ¤eld:
~ In this case the emf is
an induced electric ¤eld whose source is the changing B:
I
~ in d ¢ d~`
E=
E
Now for the static (or Coulomb) electric ¤elds we have previously discussed,
according to equation 3,
I
~ c ou lo mb ¢ d~` = 0
E
~ c ou lo mb to get
So we can add E
I ³
I
´
~
~
~
~ total ¢ d~`
E=
E in d + Ec ou lo mb ¢ d` =
E
C
(11)
C
and then Faradayzs law becomes
I
Z
d
~
~
~ ¢n
E=
E tota l ¢ d` = ¡
B
^ dA
dt
(12)
C
But here it is important to note that the electric ¤eld is measured in the rest
frame of the line segment d~`: It is the electric ¤eld felt by an electron in the
wire that would actually cause the electron to move.
It is important to note that equation (12) applies to any curve C whether
or not there is actually a wire there. It there is a conducting wire coincident
with the curve, the emf will cause a current to ¢ow. If there is no conducting
wire, the induced electric ¤eld still exists, but no current ¢ows.
2.3
Calculating induced electric ¤eld
The method for calculating induced electric ¤eld is outlined in LB §30.4. As
with the use of Gaussz Law and Amperezs law in integral form, we can only ¤nd
~ when there is su¢cient symmetry. Letzs look at an example.
the induced E
A solenoid with n turns per unit length carries a current I that is increasing
at a constant rate dI=dt: The ¤eld inside a uniform solenoid is uniform and
parallel to the solenoid axis, with magnitude B = ¹0nI: Thus as I increases,
B increases too.
dB
dI
= ¹0 n
dt
dt
Equation (12) shows that the induced ¤eld bears a similar relation to its source
(changing magnetic ¢ux) as magnetic ¤eld does to its source (current)
I
Z
~ ¢ d~` = ¹0 ~j ¢ ^n dA
B
C
9
~ in d curls around the changing ¢ux
There is a sign di¤erence, which means that E
~ in the same way
according to a left-hand rule. Thus our solenoid produces E
~ : the ¤eld lines form circles centered on
that a wire with uniform ~j produces B
the solenoid axis. If we rotate the solenoid about its axis, the picture doesnzt
~ = Eµ (s) ^µ
change, so E
Then we place a circle with radius s centered on the axis of the solenoid,
and go around it in the direction shown in (b).
I
~ ¢ d~` = 2¼sE µ
E
C
With this choice for going around C; ^n = z^; and the ¢ux is
©B = ¼s 2B
then Faradayzs law becomes
2¼sE µ = ¡¼s 2 ¹0 n
and
dI
dt
s
dI
E µ = ¡ ¹0 n
2
dt
~ is ¡^µ; as shown in the diagram. Notice that if we put a
The direction of E
wire loop in the location of our curve, current would ¢ow in the ¡^µ direction,
~ in the ¡^
and that current would produce B
z direction, thus reducing the rate
~
at which Bin sid e increases. This is required by Lenzzs law.
Something interesting happens if we use a wire loop with a small gap of
width w in it (see LB pg 970). Now current cannot ¢ow continuously because of
10
the gap. There is a burst of current as we begin to increase I; and that current
causes a build-up of charge on the surfaces of the wire, including the cut ends.
Once equilibrium is established, the net electric ¤eld inside the conducting material is zero:
~ in d + E
~ c ou l = 0
E
and thus
~ c ou l = ¡E
~ in d uc e d
E
The induced electric ¤eld is the same everywhere on the circle, but the Coulomb
¤eld changes direction in the gap, because
I
~ co u l ¢ d~` = 0
E
c irc le
Thus the total electric ¤eld is very large in the gap. Applying Faradayzs law:
¯
¯
¯ I
¯
¯Z
¯
¯
¯
¯
¯
¯
~ tota ll ¢ d~`¯ = ¼s2 ¹0 n dI = ¯
~ tota l ¢ d~`¯
E
E
¯
¯
¯
¯
dt
gap
¯
¯
circ le
'
E tota l,gap w
and
¼s2 ¹0 n dI
w
dt
Heinrich Hertz used a device like this as an antenna in his discovery of EM
waves.
Et otal,g ap =
2.4
Di¤erential form of Faradayzs law
Now we want to get the di¤erential equation that coresponds to equation (12).
I
Z
~ tota l ¢ d~` = ¡ d
~ ¢n
E=
E
B
^ dA
dt
C
11
~
We start with a curve that is at rest, so the only thing that is changing is B:
Then we can move the time derivative inside the integral on the right, and apply
Stokesz theorem to the integral on the left:
I ³
Z
´
@ ~
~
~
r £ E total ¢ ^n dA = ¡
B ¢ ^n dA
@t
C
Since this relation applies to any curve C; we must have
~ £E
~ =¡@ B
~
r
@t
(13)
This equation replaces equation (3), which is valid only for static ¤elds.
To extend this to moving curves, letzs start with a curve that has a uniform,
constant, non-relativistic velocity ~v : There is an additional contribution to the
~ Using
change in ¢ux as the curve moves to a region with a di¤ering value of B:
a Taylor series expansion, we have:
³
´
~ (~r) + ~v ¢ r
~ B±t
~ + ¢ ¢¢
B (~r + ±~r) = B (~r + ~v ±t) = B
Thus
³
´
~ (~r) = ± B
~ = ~v ¢ r
~ B
~ ±t
B (~r + ±~r) ¡ B
to ¤rst order in ±t: Thus the change in ¢ux due to motion of the curve is:
Z
Z ³
´
~
~ B
~ ¢n
±©m =
±B ¢ n
^ dA =
~v ¢ r
^ dA±t
S
and hence
S
d
©m =
dt
Z ³
´
~ B
~ ¢ ^n dA
~v ¢ r
S
due to motion of the curve. Adding the two contributions, we have
¯
Z
Z ³
´
¯
d
@ ~
¯
~ B
~ ¢n
©m ¯
=
B ¢ ^n dA +
~v ¢ r
^ dA
dt
S @t
S
tot al
and applying Faradayzs law:
½Z
I
~ 0 ¢ d~` = ¡
E
C
S
@ ~
B ¢ ^n dA +
@t
¾
Z ³
´
~ B
~ ¢ ^n dA
~v ¢ r
S
~ 0 is measured in the rest frame of d~`; i.e. the
Remember: the electric ¤eld E
~ is measured in the
frame moving with velocity ~v with respect to the lab, and B
lab frame.
Now we want to convert the last term on the right to a line integral, so we
use a result from the cover of Gri¢ths:
³
´ ³
´
³
´
³
´
³
´
~ £ B
~ £ ~v = ~v ¢ r
~ B
~¡ B
~ ¢r
~ ~v + B
~ r
~ ¢ v ¡ ~v r
~ ¢B
~
r
12
But here we have chosen ~v to be constant, and from the second Maxwell equa~ ¢B
~ = 0; so only the ¤rst term on the right is non-zero:
tion, r
³
´ ³
´
~ £ B
~ £ ~v = ~v ¢ r
~ B
~
r
Thus
I
C
~ 0 ¢ d~` =
E
=
½Z
Z h
¾
³
´i
@ ~
~
~
¡
B ¢ ^n dA +
r £ B £ ~v ¢ ^n dA
@t
Z S
Z S³
´
@ ~
~ £ ~v ¢ d~`
¡
B ¢ ^n dA ¡
B
S @t
C
where we used Stokesz theorem again in the second step. Comparing with
@ ~
equations (13), we may replace the integral of @t
B with a line integral involving
1
~
E in the lab frame :
I
I
I ³
I
´
~ 0 ¢ d~` =
~ ¢ d~` +
~ ¢ d~` =
E
E
~v £ B
f~ ¢ d~`
C
C
where again the result is true for any curve C moving at constant velocity ~v :
Thus we obtain the transformation law:
~0 = E
~ + ~v £ B
~
E
The result is consistent with the Lorentz force law, and with our previous discussion of motional emf. Thus we have established that the constant of proportionality in Faradayzs law is linked to the transformation properties of the
electric ¤eld. Wezll discuss this further later in the semester.
2.5
More on potential
Now that we have changed equation 3 to equation 13, we must rethink our ideas
~ £E
~ is no longer zero, we cannot conclude that E
~ is
about potential. Since r
the gradient of a scalar function. But equation (2) still holds, so we may still
conclude that
~ =r
~ £A
~
B
Now we insert this into Faradayzs law:
~ £E
~ =¡@ B
~ =r
~ £E
~ =¡@ r
~ £A
~ = ¡r
~ £ @A
~
r
@t
@t
@t
Thus
~ £
r
Ã
~
~ + @A
E
@t
!
=0
1 Strictly, we must apply Faradayzs law to a curve C 0 at rest in the lab that instantaneously
coincides with the moving curve C:
13
Now we can conclude that the vector
~
~ + @A
E
@t
is the gradient of a scalar function:
~
~ + @ A = ¡ rV
~
E
@t
and thus
~
~ = ¡ rV
~ ¡ @A = E
~ c ou lom b + E
~ in d u c ed
E
@t
~ by re-inserting this expression for E
~
We can con¤rm this decomposition of E
into Gaussz law:
~ ¢E
~
r
=
~ ¢
¡r2 V ¡ r
=
¡r2 V ¡
~
@A
@t
@ ~ ~
½
r ¢A =
@t
"0
~ ¢A
~ = 0; and
If we use the Coulomb gauge as we did in the static case, then r
we get
½
r2 V = ¡
"0
~
This con¤rms that the source of V is the charge density, and thus ¡ rV
is
the Coulomb ¤eld. We may not always want to use this Gauge condition in
time dependent cases, and wezll have more to say about this later, but the
decomposition still holds.
~ into induced and Coulomb ¤elds is
This mathematical decomposition of E
~ as
very useful, but of course if we put a test charge down and measure E
~
~
F =qte st wezll measure the whole E : we wonzt be able to measure any di¤erence
~ : It is important to remember that only Coulomb
between the two kinds of E
¤elds contribute to the scalar potential V :
VA ¡ V B =
Z
B
A
~ C ou lom b ¢ d~`
E
Now letzs return to motional emf and see how this plays out.
14
First look at a simple circuit with a battery and a resistor. In this circuit power
¢ows from the battery and is used in the resistor (wezll worry about how it gets
there later). There is a potential di¤erence across the resistor: ¢V = IR = E:
Now look at a similar circuit in which the power source is a person pulling
a conducting rod on conducting rails through a magnetic ¤eld:
There is a motional emf of magnitude E =`vB: But the net electric ¤eld
in the conducting rod on the right must be zero, so the magnetic force drives
charges to the sides of the circuit as shown, positive on the top and negative on
the bottom, leading to the production of a Coulomb electric ¤eld that in turn
produces a potential di¤erence across the resistor at the left. It is the electric
¤eld that drives current through the resistor.
Now we change the situation just a bit, by moving the resistance to the
15
moving rod, as shown below:
~ and it
The net force on a charge in the resistor is the magnetic force F~ = q~v £ B;
is this force that drives current in the circuit. There are no charge distributions,
no potentials, and no Coulomb electric ¤elds. The net emf is of course the same
as in the second circuit, but because the power is used in the same place as it
is produced (in the moving rod), there is no need to transfer energy to another
place in the circuit. The potential distribution in a circuit shows us how energy
is stored and redistributed throughout the system.
3
Inductance
Capacitors are devices that store energy w electric ¤eld energy w in circuits.
There are equivalent devices for storing magnetic ¤eld energy. They are inductors. Herezs how it works. Suppose we have two current loops, one carrying
current I1 and one carrying current I2 : Each loop produces magnetic ¤eld, according to the Biot-Savart law. The magnetic ¤eld produced by loop 1 threads
through both loop 1 and loop 2. B1 is proportional to I1 :
I ~
~
d`1 £ R
~ 1 = ¹0 I1
B
2
4¼
R
~ 1 through loop 2 is also proportional to I1 :
and thus the ¢ux of B
!
Z
Z ÃI ~
~
¹
d
`
£
R
1
0
~1 ¢ n
© 2 d ue to 1 =
B
^ dA2 =
I1
¢n
^ dA2 = M 21 I1
4¼
R2
S2
S2
The constant of proportionality is called the mutual inductance M 21 : We can
express it more nicely using the vector potential and Stokesz theorem:
Z
Z ³
I
´
~
~
~
~ 1 ¢ d~`2
© 2 d u e to 1 =
B1 ¢ n
^ dA2 =
r £ A1 ¢ ^n dA2 = A
(14)
S2
S2
C2
16
~:
But we already have an expression for A
~ 1 = ¹0 I 1
A
4¼
I
d~`1
R
C1
and thus
¹
M21 I1 = 0 I1
4¼
I I
d~`1 ~
¢ d` 2
R
C 2 C1
and thus
M 21
¹
= 0
4¼
I I
d~`1 ¢ d~`2
R
(15)
C2 C1
It is immediately clear from the symmetry of this expression that it doesnzt
matter which loop is labelled one and which is labelled two:
M 21 = M 12 = M
and it is also true that M is a purely geometrical property involving the size,
shape and relative position and orientation of the two loops.
Now if we change I1 ; we will change the ¢ux of magnetic ¤eld through loop
2, and so there will be an emf induced in loop 2:
¯
¯
¯
¯
¯ d©2 ¯
¯
¯
¯ = M ¯ dI1 ¯
jE2 j = ¯¯
¯
¯
dt
dt ¯
Of course the ¢ux through loop 1 also changes, and so there is also an induced
emf in loop 1.
¯
¯
¯
¯
¯ d© 1 ¯
¯
¯
¯ = L ¯ dI1 ¯
jE2 j = ¯¯
¯
¯
dt
dt ¯
where L is the self-inductance (or just inductance) of loop 1.
We can estimate the inductance of a very long solenoid of length `; area A
with N turns. The ¤eld inside is uniform and equals
B = ¹0 nI
The ¢ux through the solenoid is
© = N BA = ¹0 nN AI
and thus the inductance is
L=
©
¹ N 2A
= ¹0 nN A = 0
I
`
17
(16)
4
Magnetic energy
Since the induced emf opposes the change that creates it (Lenzzs law), the
current in a circuit cannot jump instantly to its ¤nal value: it has to build up
slowly. Letzs look at a circuit with a resistor and an inductor (coil) and a
battery with emf E0 : If we want to apply Kirchho¤ zs loop rule, we can only
use values of potential V; not induced emfs. But if we model the coil as made
of perfectly conducting wire, then the total (Coulomb plus induced) ¤eld inside
the wire is zero. Thus
~ C ou l = ¡E
~ in d u c ed
E
and integrating along the wire we get
Z b
Z b
~
~
~ in d u c ed ¢ d~`
E Co u l ¢ d` = ¡
E
a
a
or
dI
dt
To be sure we have the signs right, letzs review the physics. The induced electric
¤eld tries to oppose the increase of I in the direction of d~`; so the Coulomb ¤eld
points in the same direction as the line segment d~`; and since V decreases along
¤eld lines, the potential is higher at the end a of the coil at which the current
enters, and is lower at b where the current leaves.
Now we are ready to use the loop rule:
V a ¡ V b = ¡Eind u c e d = L
E0 ¡ IR ¡ L
dI
=0
dt
Choose a new variable x = I ¡ E0 =R: Then since E0 =R is constant,
L dx
= ¡x
R dt
which has solution
x = x 0 e¡Rt =L
At time t = 0; x = x 0 = 0 ¡ E 0=R; so
I ¡ E0 =R = ¡
and
µ
¶
E0
Rt
exp ¡
R
L
·
µ
¶¸
E0
Rt
I =
1 ¡ exp ¡
R
L
The current exponentially approaches its ¤nal value If = E0 =R: The timescale
¿ = L=R governs how fast it approaches the ¤nal value. While it theoretically
takes in¤nite time to get to If ; within 3¿ the current is within e¡ 3 = 5% of the
¤nal value.
18
1
0.8
0.6
I/If
0.4
0.2
0
1
2
Rt/L
3
4
5
Now as the current is building up, the battery is pumping energy into the circuit
at a rate
·
µ
¶¸
E2
Rt
Pb at te ry = E0 I = 0 1 ¡ exp ¡
R
L
and the resistor is using energy at a rate
·
µ
¶¸2
E2
Rt
P re sisto r = I 2 R = 0 1 ¡ exp ¡
R
L
and these two rates are not the same. After a total time T ; the energy put out
by the battery is
µ
¶¸
Z T
Z ·
E02 T
Rt
Ub atte ry =
P ba tte ry dt =
1 ¡ exp ¡
dt
R 0
L
0
"
#
µ
¶¯T
E02
L
Rt ¯¯
=
T +
exp ¡
R
R
L ¯0
½
·
µ
¶
¸¾
E02
L
RT
=
T+
exp ¡
¡1
R
R
L
while the energy used by the resistor is
Z T
Z ·
µ
¶¸2
E02 T
Rt
Ure sistor =
P res istor dt =
1 ¡ exp ¡
dt
R 0
L
0
Z
·
µ
¶
µ
¶¸
E02 T
Rt
Rt
=
1 ¡ 2 exp ¡
+ exp ¡2
dt
R 0
L
L
½
·
µ
¶
¸
·
µ
¶
¸¾
E02
L
RT
L
RT
=
T+2
exp ¡
¡1 ¡
exp ¡2
¡1
R
R
L
2R
L
So we have
Ub atte ry ¡ Ure sisto r
=
=
·
µ
¶
µ
¶¸
E02
RT
RT
L 1 ¡ 2 exp ¡
¡ exp ¡2
2R2
L
L
1 2
LI
(17)
2
19
This energy is stored in the inductor in the magnetic ¤eld. Indeed using equation (16) for the inductance of a solenoid,
¢U
=
=
=
1 ¹0 N 2 A 2
I
2
`
1 ¹0 N 2 Al 2
1
I = ¹0 n2 I 2 V
2
2
`
2
1 B2
V = uB V
2 ¹0
where V = Al is the volume of the solenoid. Thus the magnetic energy density
is
1 B2
uB =
(18)
2 ¹0
and thus the total energy ensity is
µ
¶
1
B2
u=
"0 E 2 + 0
(19)
2
¹
We can get this result more generally, starting from (17) and using (14)
I
1
1
I
~ ¢ d~`
U = LI 2 = ©I =
A
2
2
2
Now recall that we can get a more general expression by replacing Id~` with ~j d¿ :
Then
Z
1
~ ¢ ~j d¿
U =
A
(20)
2
and then from Amperezs law (4)
Z
³
´
1
~¢ r
~ £B
~ d¿
U =
A
2¹0
But
so
³
´
³
´
³
´
~ ¢ A
~ £B
~ =B
~¢ r
~ £A
~ ¡A
~¢ r
~ £B
~
r
Z h
³
´
³
´i
1
~ ¢ r
~ £A
~ ¡r
~ ¢ A
~ £B
~
B
d¿
2¹0
We use the divergence theorem on the second term to get
·Z
¸
Z ³
´
1
~ ¢B
~ d¿ ¡
~ £B
~ ¢n
U =
B
A
^ dA
2¹0 V
S
U =
The integral ¯is ¯over all space, and as long as our current is con¤ned to a ¤nite
¯ ~¯
region, then ¯A
¯ ! 0 at least as fast as 1=R2 and B goes as 1=R 3 (Remember
rule one for magnetic ¤elds- the dominant term is a dipole!) Thus the surface
integral is zero, and we have
Z
1
U =
B 2 d¿
2¹0 V
20
5
Maxwellzs equations
The equations we have derived so far are
~ ¢E
~ =
r
Coulombzs law (Gausszs Law):
~
Gausszs Law for B
:
½
"0
~ ¢B
~ =0
r
Faradayzs law
:
~ £E
~
r
Amperezs Law
:
~ £B
~
r
~
@B
=¡
@t
~
= ¹0 j
(21)
(22)
(23)
(24)
There is a nice symmetry about equations (21) and (22), once we remember
that there are no magnetic charges. But we seem to be missing something in
equation (24) because there is no time derivative. In fact we can prove that
something is missing. Take the divergence of both sides:
³
´
~ ¢ r
~ £B
~ =¹ r
~ ¢ ~j
r
0
The left hand side is zero, because the divergence of a curl is always zero. On
the right hand side, we use the charge conservation relation (5) to get
µ
¶
@½
0 = ¹0 ¡
@t
which is clearly false if @½=@t is not zero. We can see how to ¤x the problem by
taking the time derivative of equation (21):
1 @½
@ ³ ~ ~´
=
r ¢E
"0 @t
@t
~ =@t we will have a fully self-consistent set of
Thus if we add a term ¹0 "0 @ E
equations, and a nice symmetry in the two curl equations.
~
~ £B
~ = ¹0~j + ¹0 "0 @ E
r
@t
Ampere-Maxwell Law:
(25)
~ =@t is called the displacement current. Maxwell was the
The new term "0 @E
¤rst to notice the discrepancy and ¤x it, and so the law is now named for him
as well as for Ampere.
5.1
Maxwellzs equations in matter
~ = "E
~ = "0 E
~ + P~ and H
~ = B=¹
~
We have already introduced the ¤elds D
=
~
B
~
¹0 ¡ M and showed how they can simplify the static form of Maxwellzs equations
by allowing us to ignore the explicit dependence on bound charges and currents.
We obtained:
~ ¢D
~ = ½f
r
21
while the equivalent magnetic equation needs no change:
~ ¢B
~ =0
r
Faradayzs law does not involve charges and currents, so it is also unchanged:
~
~ £E
~ = ¡ @B
r
@t
~ which is related to change of bound as
But Amperezs law involves change of E;
well as free charge.
Recall that
~ ¢ P~
½b = ¡r
If we look at a tiny volume of polarized material, it will have a surface bound
charge density ¾ p = P~ ¢ ^n at each end. If we now let P~ change a little bit, the
charge on each end also increases a little bit, as if a current ¢owed from one end
to the other:
I±t = ~j B ¢ n
^ dA±t = ±¾ b dA = ±P~ ¢ n
^ dA
Thus the "polarization" current is
~
~j p = @ P
@t
This current satis¤es the same charge conservation law as regular current:
~
~ ¢ ~jp = r
~ ¢ @ P = @ (¡½b )
r
@t
@t
Thus we have four contributions to current:
conduction ("free") current due to moving charges: ~j f
magnetization current:
22
~j m ag = r
~ £M
~
~
~jp = @ P
@t
~
@E
displacement current: "0
@t
Putting all of these into Amperezs law, we have
Ã
!
~
@
E
~ £B
~ = ¹0 ~j f + ~jm ag + ~j p + "0
r
@t
Ã
!
~
~
@
P
@
E
~ £M
~ +
= ¹0 ~j f + r
+ "0
@t
@t
polarization current:
Now we combine the second term with the LHS and the third term with the
last term, to get:
Ã
!
´
~
B
@ ³~
~ £
~
~
r
¡M
= ~jf +
P + "0 E
¹0
@t
~ £H
~
r
5.2
@ ~
= ~jf + D
@t
(26)
Boundary conditions for time-dependent ¤elds
We do not need to rederive the boundary conditions for the divergence equations since they have no time-dependent terms. (Remember divergence!tunacan!bc for normal component) With ^n pointing from medium 2 into medium
1, we had
³
´
~1 ¡ D
~2 ¢ n
D
^ = ¾f
and
~ ¢ ^n is continuous
B
Now we have to look at the curl equations. Once again the rule is curl!rectangle!bc
for tangential components. Starting with Faradayzs law:
Z
Z
³
´
~
@B
~ £E
~ ¢ N^ dA = ¡
r
¢ N^ dA
@t
rec ta ng le
23
We use Stokesz theorem on the left, to get
Z
re c tan gle
³
´
~
~ ¢ d~` = ¡ E
~1 ¡ E
~ 2 ¢ ^t` = ¡ @ B ¢ N^ `w
E
@t
~
Now as we let w ! 0; the RHS! 0 because @B=@t
must remain ¤nite. Thus
we obtain the same relation as before:
³
´
~1 ¡ E
~2 £ n
E
^ is continuous
where ^t = n
^ £ N^ : Finally we look at the Ampere-Maxwell law:
Z
Z
µ
¶
³
´
~ £H
~ ¢ N^ dA =
~jf + @ D
~ ¢ N^ dA
r
@t
r ec tan g le
re c tan g le
Z
³
´
~
~1 ¡H
~ 2 ¢ ^t` = ` ~j f dw ¢ N^ ¡ @ D ¢ N^ `w
¡ H
@t
³
´
~
~1¡ H
~ 2 ¢ ^t = K
~ f ¢ N^ ¡ @D ¢ N^ w
¡ H
@t
where
~f =
K
Z
~j f dw
is the surface free current density, and again the second term on the right ! 0
as w ! 0; since the time derivative must remain ¤nite. Thus
³
´
¡
¢
~1 ¡ H
~ 2 ¢ ^t = K
~ f ¢ ^t £ ^n
¡ H
³
´
~ f £ ^n ¢ ^t
= ¡ K
So, since ^t is arbitrary,
~1 ¡ H
~2 = K
~ f £ ^n
H
24