next up previous
Next: About this document ...

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

(1)

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:

(2)


(3)

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

(4)

and

(5)



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.

(6)

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

(7)

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,

(8)

The rate at which you do work is the power you expend:

(9)

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

(10)

we see that is proportional to I1 which implies that is proportional to I1:

(11)

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
= M21 I1  
=  

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:

(12)

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 :

(13)

and

(14)

as before.

Recipe to Calculate Inductance L
Problem: Given geometry of inductor, find the self-inductance L.
Recipe for the Solution: Use

(15)

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

(16)

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:
=  
=  
B =  
=  
L =  
=  

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:

(17)

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
=  
dW = L I d I  

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

UB = W (18)

or

(19)

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.

(20)

So

(21)

To get rid of the self-inductance L, we recall that


So

(22)

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.

(23)

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

(24)

You can check this by plugging I(t) and into and seeing that it works. Let's look at

(25)

=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 ,

(26)

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

(27)

where . At t=0, . As , the current is decaying to zero as expected. The characteristic time associated with this decay is . At ,

(28)

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

(29)

The sum of the voltage drops across L1 and L2 equals the total voltage across them. So we can write


where

Leq = L1 + L2 (30)

=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,

(31)

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

(32)

=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,

(33)



 
next up previous
Next: About this document ...
Clare Yu
2002-03-05