Today in Physics 217: EMF, induction, and
Faraday’s Law
‰ Electromotive force
‰ Motional EMF
‰ The alternating-current
generator
‰ Induction, Faraday’s Law,
Lenz’s Law
‰ Faraday’s Law and
Ampère’s Law
‰ Simple example uses of
Faraday’s and Lenz’s laws
v
R
I
h
I
B
s
v×B
Box o’ B
20 November 2002
Physics 217, Fall 2002
1
Electromotive force (EMF)
A more general constitutive relation like Ohm’s Law can be
written in order to account for ways of moving charges other
than electric fields:
J =σ f .
Here f has the same units as E, but has more to it than just an
electrostatic field:
f = Estatic + fS ,
where fS represents anything else that can move charge:
chemical reactions, conveyor belts, trained bacteria,…
The line integral of f around a closed path is called the
electromotive force, E:
0
E ≡ f ⋅ dA = Estatic ⋅ dA + fS ⋅ dA = fS ⋅ dA .
v∫
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v∫
v∫
Physics 217, Fall 2002
v∫
2
“Motional” EMF
The simplest case of EMF is
provided by motion in a magnetic
v
field:
fS = × B .
c
Consider a loop of wire, partly
enclosed by a region of constant B,
moving at speed v. What’s E?
Bvh Bh ds
v

E =  × B  ⋅ dA =
=
c
c dt
c

1  dA 
= − B
 A = area filled with B
c  dt 
1 dΦ B
. Motional EMF
=−
c dt
v
R
h
v∫
20 November 2002
Physics 217, Fall 2002
B
s
v×B
Box o’ B
3
“Motional” EMF (continued)
It turns out that this last relation is valid much more
generally -- independent of the shape of the loop,
homogeneity of B -- which we will show now. It’s helpful to
keep the simpler example in mind, though.
Time: t
Consider a loop of wire
moving, and perhaps even
changing shape, through a
region with a static B, and
follow the point A. In dt it moves
a distance vdt, and with the line
element dA it sweeps out an area
da = vdt × dA
.
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t + dt
B
da
A
vdt
Border
Physics 217, Fall 2002
A
dA
4
“Motional” EMF (continued)
The change in magnetic flux through the loop, that’s
admitted by the border (shaded in cyan), is
dΦ B = Φ B ( t + dt ) − Φ B ( t ) =
∫
border
dΦ B
=
dt
v∫ B ⋅ ( v × dA )
v∫
B ⋅ da = dt B ⋅ ( v × dA ) ,
.
Now suppose a current runs in the loop. If the drift velocity
of the carriers (relative to the loop) is u, and their total
velocity w = u + v, then since u must be parallel to dA ,
v × dA = w × dA .
20 November 2002
Physics 217, Fall 2002
5
“Motional” EMF (continued)
Now,
B ⋅ ( w × dA ) = dA ⋅ ( B × w ) = − dA ⋅ ( w × B )
= −cdA ⋅ fmag
1 dΦ B
−
=
c dt
, so
v∫ fmag ⋅ dA ≡ E
1 dΦ B
E =−
c dt
,
, or
Motional EMF
same as in the simpler case.
dΦ B

In MKS: E = − dt
20 November 2002
Physics 217, Fall 2002

.

6
Example: the AC generator
Griffiths problem 7.10: A
square loop (side b) is
mounted on a vertical shaft
and rotated at angular
velocity ω. A uniform
magnetic field B is
perpendicular to the axis.
Find the EMF, and the current
driven through a resistor R in
series with the loop, for this
alternating-current generator.
ω
b
B
b
R
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Physics 217, Fall 2002
7
The AC generator (continued)
Φ B = B ⋅ a = Bb 2 cosθ = Bb 2 cos ω t
1 dΦ B ω 2
E =−
= Bb sin ω t
c dt
c
E ω Bb 2
sin ω t
I= =
R
cR
b
For MKS, leave off the factor of
c.
20 November 2002
B
Physics 217, Fall 2002
θ
ω
8
Induction and Faraday’s Law
What if the field region moves, with the loop staying still?
‰ Relativity: as long as the relative motion is the same, the
same EMF must be obtained as before. (We see it,
experimentally, to work this way, too.)
‰ In this case, though, it’s no longer clear what exerts the
force that moves the charges, since v = 0.
‰ Thus we have to postulate an induced, non-electrostatic,
electric field:
1 dΦ B Faraday’s Law
.
E = E ⋅ dA = −
(integral
form)
c dt
‰ But E ⋅ dA = ( — × E ) ⋅ da
v∫
v∫
∫
1 dΦ B
1 d
1 ∂B
B ⋅ da = − ∫
and −
=−
⋅ da ,
∫
c dt
c dt
c ∂t
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Physics 217, Fall 2002
9
Induction and Faraday’s Law (continued)
So
1 ∂B
—×E = −
c ∂t
∂B


in
MKS
.
=
−


∂t

Faraday’s
Law
‰ That is, a non-static electric field can be induced by a nonstatic magnetic field.
‰ That is, moreover, a current can be induced in a loop of
conductor by changing the flux of B through it, no matter
how the flux changes: motion of the loop, or change in B.
‰ The minus sign in Faraday’s law indicates that a changing
magnetic flux will induce an electric field and current
such that the B the current produces leads to a flux change
in the opposite direction. This is called Lenz’s Law.
20 November 2002
Physics 217, Fall 2002
10
Faraday’s Law and Ampère’s Law
Calculations of E and I with Faraday’s Law proceed just like
calculations of B from steady currents using Ampère’s Law.
Note the following forms:
4π
1 ∂B
J
—×B =
—× E = −
c
c ∂t
4π
1 dΦ B
B ⋅ dA =
E ⋅ dA = −
I enclosed
c
c dt
— × E = 0 if ρ = 0 and
—⋅B = 0
only currents change
Apparently they’re the same, with
v∫
v∫
4π
1 ∂B 4π
1 dΦ B
E ⇔ B,
J⇔−
,
.
I enclosed ⇔ −
c
c ∂t c
c dt
20 November 2002
Physics 217, Fall 2002
11
Example: polarity of motional EMF
v
Which way does the current flow
in the example of the loop being
pulled out of the region of constant B?
As we saw earlier,
Bvh
Bvh
, I=
E=
c
cR
I
h
I
.
Flux decreases as loop moves. If
current were to flow clockwise,
it would generate its own B, and
flux, that would oppose the
decrease.
20 November 2002
R
Physics 217, Fall 2002
B
s
v×B
Box o’ B
12
Example: a flux tube
Consider a cylinder, radius a, of
uniform magnetic field with timevariable amplitude B(t). (Structures
like this are seen on the surface of
the Sun.) What is the electric field?
Draw a circular loop, radius s:
v∫
1 dΦ B
E ⋅ dA = −
c dt
1 2 dB
E2π s = − π s
c
dt
1 2 dB
= − πa
c
dt
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B(t)
a
z
s dB
s dB
ˆ
φ=−
⇒ E=−
× sˆ
2c dt
2c dt
a 2 dB
⇒ E=−
× sˆ
2cs dt
Physics 217, Fall 2002
(s < a)
(s > a)
13