4Electrodynamics

IB Electromagnetism

4.2 Magnetostatic energy
Suppose that a current
I
flows along a wire
C
. From magnetostatics, we know
that this gives rise to a magnetic field B, and hence a flux Φ given by
Φ =
Z
S
B · dS,
where S is the surface bounded by C.
Definition (Inductance). The inductance of a curve C, defined as
L =
Φ
I
,
is the amount of flux it generates per unit current passing through
C
. This is a
property only of the curve C.
Inductance is something engineers care a lot about, as they need to create
real electric circuits and make things happen. However, us mathematicians find
these applications completely pointless and don’t actually care about inductance.
The only role it will play is in the proof we perform below.
Example
(The solenoid)
.
Consider a solenoid of length
`
and cross-sectional
area A (with `
A so we can ignore end effects). We know that
B = µ
0
IN,
where
N
is the number of turns of wire per unit length and
I
is the current. The
flux through a single turn (pretending it is closed) is
Φ
0
= µ
0
IN A.
So the total flux is
Φ = Φ
0
N` = µ
0
IN
2
V,
where V is the volume, A`. So
L = µ
0
N
2
V.
We can use the idea of inductance to compute the energy stored in magnetic
fields. The idea is to compute the work done in building up a current.
As we build the current, the change in current results in a change in magnetic
field. This produces an induced emf that we need work to oppose. The emf is
given by
E =
dt
= L
dI
dt
.
This opposes the change in current by Lenz’s law. In time
δt
, a charge
Iδt
flows
around C. The work done is
δW = EIδt = LI
dI
dt
δt.
So
dW
dt
= LI
dI
dt
=
1
2
L
dI
2
dt
.
So the work done to build up a current is
W =
1
2
LI
2
=
1
2
IΦ.
Note that we dropped the minus sign because we switched from talking about
the work done by the emf to the work done to oppose the emf.
This work done is identified with the energy stored in the system. Recall
that the vector potential A is given by B = × A. So
U =
1
2
I
Z
S
B · dS
=
1
2
I
Z
S
( × A) · dS
=
1
2
I
I
C
A · dr
=
1
2
Z
R
3
J · A dV
Using Maxwell’s equation × B = µ
0
J, we obtain
=
1
2µ
0
Z
( × B) · A dV
=
1
2µ
0
Z
[ · (B × A) + B · ( × A)] dV
Assuming that
B × A
vanishes sufficiently fast at infinity, the integral of the
first term vanishes. So we are left with
=
1
2µ
0
Z
B · B dV.
So
Proposition. The energy stored in a magnetic field is
U =
1
2µ
0
Z
B · B dV.
In general, the energy stored in E and B is
U =
Z
ε
0
2
E · E +
1
2µ
0
B · B
dV.
Note that while this is true, it does not follow directly from our results for pure
magnetic and pure electric fields. It is entirely plausible that when both are
present, they interact in weird ways that increases the energy stored. However,
it turns out that this does not happen, and this formula is right.