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Today in Physics 217: magnetoquasistatics
‰ Magnetoquasistatics
‰ Example
magnetoquasistatic
calculations with
Faraday’s Law
‰ Faraday-induced electric
fields and the magnetic
vector potential
‰ When does a uniformly
charged spherical shell
have nonzero E inside?
Flux tubes on the Sun, seen by NASA’s TRACE satellite.
22 November 2002
Physics 217, Fall 2002
1
Magnetoquasistatics
With Faraday’s Law, we can pose lots of new problems in
which one generates time-variable magnetic fields, and
calculates electric fields. We have derived formulas for B
generated by lots of current distributions and we can use…
‰ But can we? All those calculations assume steady currents,
and give steady magnetic fields.
‰ Thus, to use them straight away with time-varying
currents gives magnetic fields with the same time
dependence everywhere in space.
‰ This can’t be right! If I look far enough away from the
currents I should see a magnetic field that depends on
what the current was a little while ago, because the
electromagnetic information of what the current is, can
propagate to that point no faster than the speed of light.
22 November 2002
Physics 217, Fall 2002
2
Magnetoquasistatics (continued)
‰ So any calculation we do in this manner gives only an
approximate solution, good only if the current doesn’t
fluctuates very fast or if we’re not calculating the field
very far away. If τ is the time over which the current
changes and r is the distance away from the currents at
which we want to calculate the field, then we need to stay
within
Magnetoquasistatic
r cτ .
approximation
‰ You will learn how to do it right next semester in PHY
218, using retarded fields and retarded potentials. But for
now we’ll assume we’re to be working within the
quasistatic approximation.
22 November 2002
Physics 217, Fall 2002
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Magnetoquasistatics (continued)
‰ For instance, consider Example 7.9: “An infinitely long
straight wire carries a current I(t). Determine the induced
electric field, a distance s from the wire.” The result is
2 dI  s 
E=
ln   zˆ → ∞ .
s →∞
c dt  s0 
The fact that it blows up as s → ∞ is an indication of the
violation of the magnetoquasistatic approximation at
distances too great for the field to be in sync with the
current. See Example 10.2 for the sorts of things that are
necessary to get the right answer. (See it next semester,
though.)
22 November 2002
Physics 217, Fall 2002
4
Electric field from a variable solenoidal current
An infinite solenoid with radius a and n turns per unit length
carries a time-dependent current I(t) in the φ̂ direction. Find
the electric field at a distance s from the axis, inside and out.
The magnetic field is zero outside and
4π
nIzˆ
B=
c
inside, so the flux through
a circular loop with radius s
z
s is
 4π
2
nI
s
π
(s < a) a
 c
ΦB = 
 4π nIπ a2 ( s > a )
 c
22 November 2002
Physics 217, Fall 2002
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Electric field from a variable solenoidal current
(continued)
And
so
1 dΦ B
v∫ E ⋅ dA = − c dt
 4π 2 s 2 n dI
−
dt

c2
⇒ E2π s = 
2 2
 4π a n dI
−
dt
c2

(s < a)
(s > a)
 2π sn dI ˆ
− 2 dt φ ( s < a )
 c
E=
2
π
a
n dI
2
−
ˆ (s > a)
φ
 c 2 s dt
For MKS units, change one of the factors of 1/c to µ0 4π ,
and the other to unity.
22 November 2002
Physics 217, Fall 2002
6
Breaking a wire
A square conducting loop, with side a and resistance R, lies a
distance a from an infinite straight wire that carries current I.
Suppose that we break the wire, so that I drops abruptly to
zero. What total charge passes a given point in the loop
during the time this current
a
flows, and in what direction
does the induced current in
a
the square loop flow?
a
z
Field from the straight infinite
current:
2I
ˆ .
B=
φ
cs
22 November 2002
Physics 217, Fall 2002
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Breaking a wire (continued)
So the flux of B through the square loop is
2 Ia
ΦB =
c
2a
∫
a
2 Ia
ds 2 Ia
2a
ln s a =
ln 2
=
s
c
c
, and
1 dΦ B
2 a ln 2 dI
dQ
R=−
E = I loop R =
=−
dt
c dt
c 2 dt
.
Note that Q depends on the current in the loop, not I. And it’s
Q we want:
Q
∫
Q = dQ′ = −
0
2 a ln 2
2
c R
0
∫ dI ′ =
I
2 Ia ln 2
c2 R
 µ0 Ia ln 2

=
in
MKS
.


2π R


It doesn’t matter how fast the current shuts off…
22 November 2002
Physics 217, Fall 2002
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Breaking a wire (continued)
As for the direction of the induced current:
Before the wire is cut, the field points out of the page.
Shutting off the current decreases this field. The current in
the loop will flow so as to oppose this change (Lenz’s Law.) If
the induced current flows counterclockwise, it will produce a
magnetic field in the same
a
direction as the (decreasing)
field from the wire, which is
a
what it takes to oppose the
a
z
change.
22 November 2002
Physics 217, Fall 2002
9
Faraday-induced electric fields and the magnetic
vector potential
In magnetostatics,
4π
1 J ( r ′ ) × rˆ
′.
J , B= ∫
d
τ
—⋅B = 0 , —×B =
c
c
r2
For induced electric fields and ρ = 0, we can substitute
1 ∂B
,
B⇔E , J⇔−
4π ∂t
and get
1 ∂B
1 ∂ B ( r ′, t ) × rˆ
′.
, E=−
d
τ
—⋅E = 0 , —× E = −
∫
4π c ∂t
c ∂t
r2
22 November 2002
Physics 217, Fall 2002
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Faraday-induced electric fields and the magnetic
vector potential (continued)
Furthermore,
—⋅ A = 0 , —× A = B ,
4π
J.
so A depends on B in the same way that B depends on
c
ˆ
′
B
r
,
×
t
r
( )
1
Thus
A=
dτ ′ , and
2
4π
r
1 ∂A
E=−
.
c ∂t
∫
1 ∂
1 ∂B
Check: — × E = −
—× A = −
c ∂t
c ∂t
(Faraday's Law again).
Leave off the factor of 1/c for MKS.
22 November 2002
Physics 217, Fall 2002
11
When does a uniformly charged spherical shell
have nonzero E inside?
A spherical shell with radius R carries a uniform charge
density σ. It spins about the z axis at angular velocity ω ( t )
that changes with time. Find the electric field inside and
outside the sphere.
The week before last (Example 5.11, lecture, 11 November
2002), we worked out the magnetic vector potential from a
spinning, uniformly charged spherical shell, and got
 4π Rσω
ˆ
 3c r sin θ φ ( r < R ) ,

A(r ) = 
4
4
R
π
σω sin θ

ˆ (r > R).
φ
3

3c
r
A is time dependent if ω is.
22 November 2002
Physics 217, Fall 2002
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When does a uniformly charged spherical shell
have nonzero E inside? (continued)
So there are two contributions to the electric field:
electrostatic, and induced. The static part is, of course,
0 ( r < R ) ,

Estatic ( r ) =  4π R 2σ
ˆ (r > R).
r

 r2
Using the E-A relation we just derived, the induced part is
 4π Rσ dω
ˆ (r < R) ,
r sin θ φ
−
2 dt
1 ∂A  3c
Eind ( r , t ) = −
=
c ∂t  4π R 4σ dω sin θ
ˆ (r > R).
φ
−

3c 2 dt r 3
22 November 2002
Physics 217, Fall 2002
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When does a uniformly charged spherical shell
have nonzero E inside? (continued)
So the total electric field is
 4π Rσ dω
ˆ (r < R)
φ
−
θ
sin
r

2 dt
 3c
E (r ,t) = 
, or
2
4
 4π R σ rˆ − 4π R σ dω sin θ φ
ˆ (r > R)
 r 2
3c 2 dt r 3
 µ0 Rσ dω

ˆ
− 3 dt r sin θ φ ( r < R )



= 2
 in MKS.
4
 R σ rˆ − µ0 R σ dω sin θ φ
ˆ ( r > R )
 ε 0 r 2

dt r 3
3
It’s nonzero inside, if the angular velocity changes with time.
22 November 2002
Physics 217, Fall 2002
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