Dr. Carolyn Morgan's Thursday quiz section is regularly attended by one student only. It is planned to cancel this section. Anyone currently enrolled in the Thursday quiz section should choose one of the 3 remaining sections: Tuesday 10:40 -11:35, Wednesday 11:45 - 12:40, or Friday 11:45 - 12:40. Inform Dr. Morgan of your choice so she can pass your quiz grade(s) to the appropriate instructor. If none of these choices is possible for you, inform Dr. Morgan and myself.
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
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:
If the conducting circuit has total resistance, R, then the current is
Faraday's law doesn't give us power for free. If the induced emf is used to create a current in a circuit, then the power supplied to the circuit must come from somewhere. Where? It comes from the power that must be expended to move the bar through the magnetic field. The power expended in the circuit is P = I emf = I B l v. By equating this power to the power expended in moving the bar, we can determine the force required to move the bar. The work done by an applied force, Fapp, in moving the bar a distance Dx is W = Fapp Dx. If this takes time Dt, the power delivered by the applied force is P = W / Dt = Fapp Dx / Dt = Fapp v. Equating the power delivered with the power expended gives:
This force is the same magnitude as the force exerted on the bar by the magnetic field due to the current through the bar:
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:
Application of Lenz's law will tell us the direction of induced currents, the direction of applied or produced forces, and the polarity of induced emf's. The idea is to determine the direction of change of the magnetic flux, then find the direction of induced current that will produce magnetic flux in the opposite direction. Up to now we haven't really spoken of a direction for magnetic flux, but since it is derived from the magnetic field lines, and they have a direction, we can use the direction of the lines. If the magnitude of the flux is increasing, then the change is in the direction of the magnetic field (the part going through the loop). If the magnitude of the flux is decreasing, then the change is opposite the direction of the magnetic field.
For instance, consider again the bar moving along rails through a magnetic field. As the bar moves to the right, more magnetic field lines pass through the loop, so the magnitude of the flux is increasing. We say it is increasing into the page (the magnetic field is into the page).
Lenz's law says that the induced current will produce magnetic flux opposing this change. To oppose an increase into the page, it generates magnetic field which points out of the page, at least in the interior of the loop. Such a magnetic field is produced by a counterclockwise current (use the right hand rule to verify).
If instead the bar is moving to the left, then the magnitude of the flux is decreasing as the area of the loop decreases. The change of flux is directed opposite the magnetic field, or out of the page. The induced current will produce a magnetic field opposing the change, in this case a magnetic field directed into the page. And using the right hand rule, this requires a clockwise current. We reverse the motion of the bar and the direction of the induced current is also reversed.
In Figure P20.23, what is the direction of the current induced in the resistor at the instant the switch is closed.
To solve problems of this type, involving Lenz's law, doesn't require calculations, but it does require an understanding of the right hand rule, and careful consideration of the direction of change of magnetic flux. In this problem, before the switch is closed, there is no current in the solenoid, and no magnetic field inside. After the switch is closed, current will flow and a magnetic field is created in the solenoid (you can assume that the field outside of the solenoid is zero). The loop containing the resistor encloses the solenoid, so the magnetic field will pass through the loop, that is, there is magnetic flux through the loop after the switch is closed. The change from zero flux to non-zero flux will produce a current in the loop and passing through the resistor. Now we must determine the direction of that current.
First, let's find the direction of the current through the solenoid. The current will flow from the positive battery terminal to the negative, so it will flow from the battery, through the switch, then through the solenoid, and back to the battery. The resulting magnetic field is directed to the left. The induced current in the loop with resistor will produce a magnetic field directed to the right. That means the current through the resistor will flow left to right.
Notice that if we now ask for the direction of the current when the switch is opened again, we will get the opposite answer. That is, the magnetic field of the solenoid is initially pointing to the left, and goes to zero when the switch is opened. The magnitude of the magnetic flux is decreasing, so the change in the flux is opposite the direction of the field (the field that exists during the change). The change in the flux is directed to the right. The induced current in the loop with resistor will produce a magnetic field directed to the left, and therefore the current through the resistor is from right to left.
While the switch is not moved, remaining either opened or closed, the magnetic flux is not changing, and there is no current in the loop with resistor.