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Recall from last lecture:

Ch. 20: Induced Voltages and Inductance

We have seen that a magnetic field exerts a force on a wire carrying a current, and that a wire carrying a current generates a magnetic field. Currents are produced by electric fields, so there seems to be some connection between electricity and magnetism. In this chapter, we make that connection, seeing how a magnetic field can produce a potential difference. Magnetic flux will play an important role throughout this chapter.

20.1 Induced emf and Magnetic Flux

Experiments in the 19th century showed that a changing magnetic field can produce an emf. We quantify the change in terms of magnetic flux. Magnetic flux is defined in a similar manner to electric flux. For a loop of wire with area A, in a magnetic field, B, the magnetic flux, F is given by:

F = Bperp A = B A cosq
where q is the angle between the perpendicular to the plane of the loop, and the magnetic field B. The SI units of magnetic flux are Tm².

Another way of looking at flux is to think of it as a count of the magnetic field lines which pass through the loop. If the loop is oriented perpendicular to the field (q=0) then the flux will be large. If the loop is oriented parallel to the field (q = 90°) no magnetic field lines pass through the loop, and the flux is zero.

Example: P20.2

A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the field makes an angle of 50.0° with the normal to the plane of the loop, as in Figure 20.2, determine the magnetic flux through the loop.

From what we are given, we use
F = B A cosq = (0.300T)(2.00m)²cos50.0° = 0.386 Tm²

20.2 Faraday's Law of Induction

Faraday's law of induction relates the change of magnetic flux to induced emf:

emf = -N (DF / Dt)
where N is the number of loops in the coil where the emf is induced, and DF is the change in flux that occurs in time Dt. This is one of the instances where the term emf differs somewhat from voltage, and therefore I will use it.

The induced emf is proportional to the change of magnetic flux, F = BAcosq. There are three ways in which the flux can change:

  1. a change of the magnitude of the magnetic field, B,
  2. a change of the angle between the loop and the magnetic field, q, or
  3. a change of the area of the loop, A.

The minus sign is there to remind you that the polarity of the induced emf opposes the change of flux. This is stated precisely in Lenz's law:

The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop. That is, the induced current tends to maintain the original flux through the circuit.

Example: P20.13

A wire loop of radius 0.30m lies so that an external magnetic field of strength +0.30T is perpendicular to the loop. The field changes to -0.20T in 1.5s. (The plus and minus signs here refer to opposite directions through the loop.) Find the magnitude of the average induced emf in the loop during this time.

The loop is always perpendicular to the field, so the normal to the loop is parallel to the field, so q = 0, and cosq = 1. The flux through the loop is therefore F = BA = Bpr². Initially the flux is
Fi = (0.30T)p(0.30m)2 = 0.085Tm²
and after the field changes the flux is
Ff = (-0.20T)p(0.30m)2 = -0.057Tm²
The magnitude of the average induced emf is:
emf = DF/Dt = (Fi - Ff)/Dt = (0.085T - (-0.057T))/1.5s = 0.095V = 95mV.

Example: P20.10

The flexible loop in Figure P20.10 has a radius of 12cm and is in a magnetic field of strength 0.15T. The loop is grasped at points A and B and stretched until it closes. If it takes 0.20s to close the loop, find the magnitude of the average induced emf in it during this time.

This is a case where the change in flux is caused by a change in the area of the loop. Both the magnetic field and the angle q remain constant. When the loop is stretched so that its area is zero, the flux through the loop is zero. The change in the flux is thus equal to its original value,
Fi = B A cosq = (0.15T)p(0.12m)² = 6.8×10-3Tm²
The average induced emf is thus:
emf = N (DF / Dt) = (6.8×10-3Tm²)/(0.20s) = 3.4×10-2V = 34mV.

20.3 Motional emf

An interesting application of Faraday's law is to produce an emf via motion of the conductor. As a simple example, let's consider a conducting bar moving perpendicular to a uniform magnetic field with constant velocity v. For this first look, we have just a bar, not a complete conducting loop, and we will consider what happens using just the force on a moving charge, F = qvBsinq. This force will act on free charges in the conductor. It will tend to move negative charge to one end, and leave the other end of the bar with a net positive charge.

The separated charges will create an electric field which will tend to pull the charges back together. When equilibrium exists, the magnetic force, F=qvB, will balance the electric force, F=qE, such that a free charge in the bar will feel no net force. Thus, at equilibrium, E = vB. The potential difference across the ends of the bar is given by DV = E l, or

DV = E l = B l v
This potential difference exists because of the excess of charge at the ends of the conductor created by motion through the magnetic field. If the direction of motion is reversed, so is the polarity of the potential difference.

Now let's consider what happens when we add conducting rails for the top and bottom of the bar to contact, and a resistor between the rails to complete a loop. We can apply Faraday's law to the complete loop. The change of flux through the loop is proportional to the change of area from the motion of the bar:

DF = B DA = B l Dx.
Using Faraday's law, we find the magnitude of the emf to be (N = 1):
emf = DF/Dt = B l Dx/Dt = B l v
where I have used the relation v = Dx/Dt. This is the same result we obtained considering the conducting bar by itself.

If the conducting circuit has total resistance, R, then the current is

I = emf / R = B l v / R

Example: P20.18

Over a region where the vertical component of the Earth's magnetic field is 40.0µT directed downward, a 5.00 m length of wire is held in an east-west direction and moved horizontally to the north with a speed of 10.0 m/s. Calculate the potential difference between the ends of the wire, and determine which end is positive.

The vertical component of the magnetic field is perpendicular to the wire and its motion, so this is what we need. Using the expression obtained for DV yields:

DV = B l v = (40.0 µT)(5.00 m)(10.0 m/s) = 2.00 mV
To determine which end is positive, consider a positive charge moving north through a downward magnetic field. The right hand rule gives a force directed to the west. So the west end of the wire will have a net positive charge, and a more positive potential.

© Robert Harr 2000