Last Name: 
CID:           

Note: Items in red are measurements that must be made as you do the experiment.

Introduction You will study Faraday’s Law by observing the voltage produced by a bar magnet passing through a coil of wire. Your experimental apparatus consists of a simple bar magnet that moves slowly through a liquid-filled tube wrapped with an 8000-turn coil of wire. When the tube is turned upside-down, the magnet falls through the coil. In this process, the magnetic flux passing through  the coil changes and a voltage, or emf, is induced across the coil.  This emf is measured with a computer-monitored voltmeter and plotted on the computer as a function of time.
Lenz's Law The magnet inside the tube has blue tape on one end and white tape on the other. Let the magnet fall through the coil both ways. By using Lenz's Law, determine which direction current must flow in order to oppose change in the magnetic flux. (Since you don't know which end of the magnet is the north pole, make a guess and see if your prediction matches the results.)  The voltmeter resisters the voltage as positive when current is flowing through the loop in the direction marked by the arrow. The north pole is: Select blue or white: Blue White
Faraday's Law Faraday’s Law, Eq.1, relates the emf in the coil to the rate that the flux through the coil changes in time.  By integrating Faraday's Law, we can find the total change in flux, ΔΦ (Delta Phi), through the coil. Since flux represents the number of magnetic field lines passing through the coil, ΔΦ (Delta Phi) is just the final number of field lines minus the initial number of field lines passing through the coil. The computer produces a plot of the emf as a function of time, so the integral in Eq. 2 is just the total area under the curve. When the magnet is far from the coil, the flux through the coil is negligibly small. When the magnet reaches the midpoint of the coil, all of the field lines leaving the north pole (or coming into the south pole) of the magnet pass through the coil. At this point the flux has reached its maximum value. For an instant, the flux through the coil remains constant, so no emf is produced, and the graph crosses the x-axis. (Remember that the flux is the integral of the emf curve, so be careful not to confuse the point where the flux is largest and the point where the emf is largest!)  Then as the magnet continues to fall through the loop, the emf reverses and the the total change in flux  - the area under the curve - decreases until it eventually becomes zero. (Initially there was no flux through the coil. At the end of the experiment, there was no flux through the coil. The net change in number of field lines is then 0-0=0, so ΔΦ(Delta Phi=0)). Eq. 9.1     Eq. 9.2 1.      Obtain a printout of ε(t), the emf as a function of time. Note the position on the graph that corresponds to the midpoint of the coil.  The emf will pass rapidly through zero at this position.   Manually integrate the number of grid squares under the curve prior to this point.         Number of squares =  2.      Calculate the change in magnetic flux represented by one grid square: Eq. 9.3       In this equation Δε (Delta emf) represents the distance between lines on the y axis in volts, and Δt (Delta t) represents the distance between lines on the x axis in seconds. (We ignore the minus sign, as it only indicates direction.)          Change in flux for one square = Wb 4.      Multiply this value by the number of squares obtained in Step 1 to get the total magnetic flux of the magnet. (The change in flux to the point that the magnet is midway in the coil is total flux of the magnet.)         Total flux of the magnet: Wb 5.   Assuming that the magnetic field inside the magnet is uniformly distributed over its 5mm radius, calculate the magnitude of the magnetic field inside the magnet. (Remember that flux is BA.)          Magnetic field strength: T