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Lectures on Chapter 23: Faraday's Law and Inductance
EMF
It takes work to move charges around a circuit. The emf
is the amount of work done per unit charge
Recall that work is defined as
The work done in dragging a charge around a circuit is
Let be the force per unit charge
Then the work per unit charge done in dragging a charge around a
circuit is the emf :
We might be tempted to say that
where is the electrostatic electric field set up by
charges. But you can't get a steady current flow by putting some
charges near a wire. If you put a wire in an external electrostatic
field produced by stationary charges, the charges in the
conducting wire would quickly rearrange themselves to make the wire
an equipotential. But you need a voltage drop to get current to flow.
Another way to see that
is to note that
Emf due to
is zero.
Recall our definition of electric potential
=1.0 true in
circle_a.eps
If we start at point , go around a loop, and end at point
a, then
So in our expression
,
is a force per unit charge that is not due to electrostatic
charges. is a force that keeps the charges moving.
doesn't do this. can be the electric
field produced by an ongoing chemical reaction in a battery.
Or can be the force produced by the conveyor belt in a van de
Graaff generator.
One of the most important sources of an emf is
caused by the motion of a loop of wire through a
magnetic field. This is the principle behind
electric generators. In the shaded region,
there is a uniform magnetic field ,
pointing into the page, and the resistor R
represents whatever it is (maybe a light bulb
or a toaster) we're trying to drive a current
through.
=3.0 true in
moving_loop.eps
If the loop is pulled to the right
with velocity , the charges in the vertical
segment a b feel a magnetic force qvB in the
direction of the wire. (The other segments feel forces
perpendicular to the wire.) The figure below shows the forces
if I = 0.
=3.0 true in
moving_loop2.eps
The force qvB on the charges in the segment ab drives a current
around the loop in the clockwise direction. The force per unit
charge is
=2.0 true in
loop.eps
Let h = height of loop = distance from a to b. Then the emf is
The integrals from
and from
are 0 because
. The integral
from
is zero because it is outside the field.
So
We can relate the emf to the magnetic flux through the loop.
This is like the electric flux
that we defined for Gauss' law. However, the surface S we are
integrating over for is not closed. is an element of
the surface pointing in the direction normal to the element da.
Which normal you pick is arbitrary at the moment. But it will turn
out that points according to the right hand rule for the curve C.
. You can think of as the
number of magnetic field lines piercing the surface
S.
=1.5 true in
loop_Bfield.eps
Units
Let's calculate the flux through our loop.
=2.5 true in
moving_loop_hx.eps
Let's choose the normal to the loop
to the point into the page.
When we evaluated
, we
went around the loop in a specific direction.
If the fingers of our right hand curl in this
direction, our thumb points in the direction
of the normal . So points in the same direction as
. So points in the same direction as . Then
As the loop moves, the flux decreases:
We need the minus sign to indicate that the flux is decreasing.
Compare this with our expression for the emf:
This is called Faraday's Law.
Thus the emf generated is minus the rate of change of flux through
the loop. We derived this relation for a special case but it is true
in general. This is true for a loop of arbitrary shape moving in any
direction through a magnetic field. If we have a coil with N turns
of wire that experiences changing flux, then an induced emf appears
in every turn and we add these emf's:
Faraday's law is the principle behind electric generators. Emf that
can drive electric currents are generated by changing the flux
through coils of wire.
=2.5 true in
moving_loop_negX.eps
So far we have been considering a static field and a moving
loop. What happens if we hold the loop stationary and move the
source of the magnetic field with a velocity opposite to that of the
loop when it was moved? Faraday found that exactly the same emf is
produced. What matters if the relative motion of the loop and
. How did Faraday interpret this? What makes the charges
move?
because the velocity of the charge is zero (the loop is
stationary). To get the electrons to flow, we need an electric
field. This must mean that a time varying field
produces an field! To see this mathematically, note that
and
Here the closed loop C is spanned by the
surface S.
=1.5 true in
circular_loop.eps
The direction of around
C is related to the direction of by
the right hand rule. Notice that the changing
magnetic field on the right hand side of the
equation produces the field on the left hand side.
Faraday's law tells us that we can generate an field simply by
changing the magnetic flux. It doesn't matter how we change the
flux. We can move the loop, or we can move the magnet, or we can
just increase or decrease the magnetic field while the loop and the
magnet are stationary. Actually, you don't need to have a wire to
induce an field. In empty space, a changing field
will induce an field. Applications: generators, microphones,
stereo speakers.
Lenz's Law
Lenz's law gives us an early way to keep track of the direction of
current flow induced by changing flux:
``The emf induced is in such a direction, that if a current flows, the
magnetic field produced by the induced current opposes the change in
flux that produced the emf."
This is the meaning of the minus sign in Faraday's law. Analogy:
Let's suppose money going into your bank account is like magnetic
flux going through a loop of wire. Suppose you earn $1000/month
which you put into your bank account. Now if your boss says that
you're going to get a pay cut and that you'll be earning $500/month,
you would oppose this change. There's still money going into your
bank account, but it's less. There's still flux going through the
loop, but it's less. So you would try to supplement your earnings
with an extra job. The loop has an induced current flowing in it
that supplements the external reduced flux.
=3.0 true in
lenz_law.eps
You probably wouldn't oppose a salary raise so let me draw another
analogy. Let's suppose you are in a boat in the middle of a lake.
The boat springs a leak and water starts to pour in. This is like
increasing flux through the loop. You would oppose this change by
trying to reduce the amount of water pouring in; i.e., by bailing
water out of the boat. Similarly, the induced current in the loop
produces flux opposite to the increasing external
flux.
=2.5 true in
increasingB_lenz.eps
The basic message of Lenz's law: Oppose change.
Example: Jumping Ring
=1.0 true in
jumping_ring.eps
If you place a metal ring on top of a solenoid,
then turn on the current in the solenoid, the
ring will jump off. (It helps to have an iron
core in the solenoid to increase the field.)
Why does the ring jump off? When you turn on the
current, the flux through the ring suddenly increases from zero to
non-zero value
. This induces
an emf in the ring that drives a current in the ring. According to
Lenz's law, the induced current in the ring will produce a
field in the opposite direction to that of the solenoid. So like 2
bar magnets
, the ring and solenoid repel and
the ring flies off.
Example
=3.0 true in
EB_field.eps
A uniform magnetic field
, pointing straight up, fills
the shaded circular region. If it changes with time, what is the
induced electric field ?
Solution: Draw a loop C of radius r and apply Faraday's law.
where S = surface spanned by loop C. Let's evaluate the
left hand and right hand sides of this equation.
Once you figure out which way
points, use the right
hand rule to get the direction of .
=2.0 true in
cylinder_B.eps
If the circular region is
mounted on a wheel that's free to rotate and there is a line of
uniform charge mounted on the rim of the wheel, then the
wheel will start to spin as the field
decreases because the induced field will push
the charges and hence the wheel. Where does
the angular momentum come from? It can't
come from nowhere because angular momentum is
conserved. It turns out that the angular
momentum comes from
the and fields. and have energy,
momentum, and angular momentum. The angular momentum density is
.
Power Dissipation
=3.0 true in
moving_loop_force.eps
Let's revisit the loop of wire that we were pulling out of the
magnetic field. We saw that the emf induces a current I that tries
to replenish the flux. The current in the loop feels a force
The forces are shown in the figure.
Since
, they cancel out. Only
survives. opposes the force we use to pull
the loop out of the field. Note that
and
. So
We can find I by noting that
and
You do work on the loop because opposes you. You must
apply at least a force that is equal and opposite to
to keep the loop moving. is like friction
force. (Do you need to apply more than to keep the loop
moving? No, if = constant,
The rate at which you do work is the power you expend:
The rate at which you do work is turned into power dissipated in the
resistor R:
Notice that I2R = Fv, i.e., the rate of dissipating power in the
resistor equals the rate at which you do work.
Inductance
Suppose we have 2 loops of wire with one above the other.
=2.0 true in
twoLoops_Bfield.eps
If we run a current
I1 through loop 1, it will produce a magnetic field
. Let
be the flux of
passing through loop 2:
From the Biot-Savart law
we see that
is proportional to I1 which implies that is
proportional to I1:
where the constant of proportionality M21 is called the
mutual inductance.
There are 2 things to note about mutual inductance:
- 1.
- This is a purely geometric quantity. It depends on the sizes,
shapes, and
relative distance of the loops. It does not depend on the current
I1.
- 2.
-
M21 = M12 = M.
This implies that the flux
through 2 produced by a current Iflowing through loop 1 is exactly the same as the flux ,
through 1 produced by I flowing through loop 2:
=1.0 true in
twoLoops.eps
If we vary the current in loop 1, will vary and this will produce an
emf in loop 2:
So loop 1 can generate a current in loop 2 without touching it.
Note that if the flux is through a coil with N turns, then
.
=2.0 true in
coil_Bfield.eps
Thus
Self-Inductance
We don't need two loops to see this effect. If we have only one loop
with a current flowing around it, the current produces a
field. This field produces a flux through the loop.
=1.0 true in
Bfield.eps
The flux is proportional to the current:
where I is the current. The constant of proportionality L is called the
self-inductance of
the loop. As with M, it depends solely on the geometry (size and shape) of
the loop. If the current changes, then there will be a changing flux
through the loop which in turn will produce an emf in the loop given
by Faraday's law:
The minus sign implies that whenever we change the current, the change produces
a "back emf" that opposes the change. Thus currents want to stay constant. L
gives the system inertia. It acts like a mass does in a mechanical system.
Units of Inductance
Inductance is measured in henries (H):
Note that if we replace a loop with a coil that has N turns, then
:
and
as before.
Recipe to Calculate Inductance L
Problem: Given geometry of inductor, find the self-inductance L.
Recipe for the Solution: Use
- 1.
- Assume current I flows through the inductor.
- 2.
- Calculate
using Ampere's Law or Biot-Savart Law.
- 3.
- Calculate flux
.
- 4.
- Use
where N = total number of turns of
wire in the inductor
Inductance of a Solenoid
=2.0 true in
solenoid.eps
Suppose we have a very long solenoid with cross sectional area A and
n turns per unit length. What is the self-inductance per unit length
?
Solution: Use
For a length of the solenoid, there are turns.
Assume a current I flows through the coil.
(This is analogous to assuming a charge q on a
capacitor when we calculate capacitance.) To calculate the flux
, we need .
=2.0 true in
ampere.eps
To find , use Ampere's law:
Energy Stored in Magnetic Fields
In order to get current started in a current loop, we need to do work against
the back emf that opposes the increase in current. The work we do gets stored
as potential energy. So to find the potential energy, we calculate how much
work is done as follows. The rate of doing work (or power going into the
inductor) is given by:
Note that dq/dt = I, the current. To find dW/dq, note that
the back emf
. So the work done by the
battery per unit
charge to overcome the back emf is
.
Recall that
is the work lost per unit charge in going
through the inductor. The energy gained by the magnetic field is minus this.
Thus
Integrate, starting from I=0 at t=0, to get the total amount of
work done by the battery to get to current I:
Since this work goes into magnetic potential energy
UB, we have
or
Once we get the current going, where did all the energy we put into the system
go? Answer: the energy is stored in the magnetic field. In fact the energy
density is given by
This equation is true in general. However, let's prove it for the special case
of an infinite solenoid.
=2.0 true in
solenoidSection.eps
Plug
into
where is the length of a section of the solenoid and A is the cross
sectional area of the solenoid. So is the volume of that
section of solenoid. To get uB in terms of B, note that
in a solenoid.
So
To get rid of the self-inductance L, we recall that
So
as advertised.
This is the magnetic analog of
.
RL Circuits
Consider a circuit with a resistor and an inductor in series.
=2.5 true in
RLcircuit.eps
When we close the
switch on
, current starts to flow
. The rise in current produces an induced emf in the inductor that
opposes the rise in current:
. As time
goes on, the rate of increase of the current becomes less rapid and the
magnitude of the self-induced emf,
, decreases
because dI/dt decreases. As
, the current
becomes steady and constant,
, and
, i.e. the total
voltage drop is across the resistor.
When the switch
is closed, we get the equivalent
circuit:
=2.5 true in
RLcircuit2.eps
Let's apply the loop rule. We go around the circuit in a clockwise
fashion. Then we get voltage drops across the resistor and inductor
and a voltage gain from the battery.
or
We want to solve this for I(t) with the initial condition
I (t=0)
=0, i.e. initially there is no current. The solution is
You can check this by plugging I(t) and
into
and seeing that it works. Let's look at
=2.0 true in
currentResistor.eps
At t=0,
as desired. Since the voltage drop
across the resistor is IR, there is no voltage drop across the resistor
at t=0. All the voltage drop is initially across the
inductor. As
,
, and
, i.e., all the voltage drop is across the
resistor. Since I approaches a constant
,
and no voltage drops across the inductor
as
. Graphically we have:
=4.0 true in
VRVL.eps
Time Constant
If we write
where
,
the inductive time constant
describes the
characteristic time over which the current rises to an appreciable value.
(Since the exponent -Rt/L must be dimensionless, L/R must
have the dimensions of time. You can also check explicitly that
[L/R] = time.) If we set
,
i.e., the current has reached 63% of its final value at
.
=2.5 true in
RLcircuit.eps
What happens if we remove the battery by closing the switch to b?
(Close bbefore opening a so the current keeps flowing.) Then our circuit
looks like
=2.0 true in
RLclosedCircuit.eps
Without a battery the current will decrease with time: dI/dt < 0.
Once again the inductor will resist this change by producing a back
emf (-LdI/dt). We can write down a differential equation for
this circuit by
applying the loop rule (or by setting
in our previous
equation):
The solution of this equation is
where
.
At t=0,
.
As
,
the current
is decaying to zero as expected. The characteristic time associated
with this decay is
. At
,
i.e. the current has decayed to 37% of its original value.
=2.0 true in
currentDecay.eps
Adding Inductances
(Your book forgets to tell you this.)
If there are 2 or more inductors in a circuit you add them the same way as you
do resistors.
In Series
=2.0 true in
InductanceSeries.eps
Suppose we shut switch S in the circuit shown. Then the current
I starts to
flow. Since I increases,
. The loop rule gives
The sum of the voltage drops across L1 and L2 equals the
total voltage across them. So we can write
where
=2.0 true in
InductanceEq.eps
Compare this to
Req=R1+R2 which is the relation for
resistors in series. In general, for n inductors in series,
In Parallel
=2.0 true in
InductanceParallel.eps
Suppose we shut the switch in a circuit where the inductors are in parallel.
Then the voltage drop across each inductor is the same but the current through
each is different if
. Thus
Now use the junction rule:
where
=2.0 true in
InductanceEq.eps
Compare this to
1/Req = 1/R1 + 1/R2 for resistors in
parallel. In general, for n inductors in parallel,
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Clare Yu
2002-03-05