The goals of this experiment are to:
- observe the effect of a magnetic field on a charged particle.
- calculate the charge to mass ratio of an electron
- verify the "right hand rule" which governs the direction of the force on the charge.
- determine the magnetic field of the Earth and a solenoid using a Gaussmeter
Notes:
We have talked about the movement of an electron in an electric field. It is only sensible, therefore, to discuss the movement of an electron in a magnetic field, now that you have (hopefully) discussed magnetism in your lecture. Here's a brief recap:
If a charged particle is moving in a magnetic field (B), then it experiences a force of magnitude:|F| = q |v x B| = qvBsin(theta) where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and theta is the angle between v and B. The direction of the force is easily obtained using the right hand rule. (Shown in class!)Notice the fact that the magnetic force does no work on the charged particle. That means that the magnetic field cannot alter the speed of the particle. It can only change the particle's direction. A charged particle moving perpendicular to a magnetic field will move in a circle. You may wish to refer to this applet to view the effects of the magnetic field on a charged particle.
In our experiment today, we will use the cathode ray tube (CRT) again to view the effect of a magnetic field on a charged particle. We will place the CRT in a magnetic field produced by a solenoid (a coil), and observe what happens to the image on the screen.
A solenoid is a coil of wires. It produces a magnetic field inside it with a magnitude of B = (mu)* nI, where mu is the magnetic permeability of free space, n is the number of turns per unit length, and I is the current going through the coil. For a particle moving perpendicular to this field with a speed v, the radius of the circular path it takes will be:
R = mv /qB If we know the radius of the circular path and the magnetic field strength B, then we can figure out the value m/q! That means that all we have to do to figure out how much mass an electron has is place it in a magnetic field and watch what path it takes! (We can figure out how much charge it has by other means.) We can find the mass of any charged particle using this method. In this experiment, we will be finding the ratio of charge to mass, or q/m, of an electron. After you complete this lab, you will be able to brag that you have found the mass of an electron without having to go look it up in a book. : )Actually doing the experiment with our CRT is just a little more complicated. We could observe the electron's path with a bubble chamber or a cloud chamber, and the calculation would be much simpler, but unfortunately, getting these apparatus to work takes a little doing (believe me!), and requires special cameras, liquid nitrogen or dry ice, and Mel (our shop guy) would probably kill me if I broke any **more** of them! So with a few more steps we can calculate what will happen in our CRT.
First, let's find the period of the circular movement of the electron.
T = 2 * pi * m / q* B No matter what the speed of the particle, the period is always this value.
Now let's consider the fact that our electron is not moving perpendicular to the magnetic field. It has some component of the velocity in the x direction (where the x direction is the same as in the electric deflection lab). It will not move in a circle anymore. It will have a helical path around the x axis. However, if I look down the x axis, I will see the electron appear to move in a circle. Just think of the path of the electron as the sum of these two movements: a circular movement caused by the magnetic field, and a linear movement caused by the initial velocity in the x direction. Neither movement affects the other.
Therefore, the equation of motion in the x direction is just x = v_ox * t. If we take the time t to be the period of one circlular movement, then the distance travelled is:x = v_ox * T = v_ox * 2 * pi * m / q * B We also know that the initial velocity in the x direction is caused by the accelerating voltage, V_g, from the electric deflection lab (v_ox) ^ 2 = 2 * q * V_g / m. If we can find the magnetic field at which the electrons hit the screen after one period, we can find the quantity q/m.
Part I
1. Measure V_g. You will need to exercise the same caution that you used in the previous (electric deflection) lab. (Never connect or disconnect any wires while the power supply is on!).2. Count the number of turns (loops) in the solenoid, and measure its length. n = # of turns / length
3. Place the solenoid over the CRT.
4. Plug in the large power supply into the solenoid.
Be cautious! The power supply puts out a great deal of current (>10 A!). Do NOT use the multimeter to measure this current. Turn the power supply on, making sure that you are not touching any of the connections.
5. Determine the current at which the image on the CRT becomes a point. This corresponds to the case at which the electron has gone through one period. ***To determine the current, read the dial on the solenoid's power supply***. As soon as you finish this step, turn off the power supply to the solenoid.
6. Calculate q/m.
Part II
Verify the Right Hand Rule:
1. Obtain a bar magnet.
2. Move one side of the magnet (N or S) close to the CRT.
3. Write your observations in your data section.
4. Repeat for the other side of the magnet.
Part III:
1. Obtain a Gaussmeter. (If none are available, please be patient. You may not join another group, since their readings will be slightly different from yours.)
2. Connect a multimeter to the Gaussmeter.
3. Turn the multimeter to DCV or the V with a bar over it.
4. Turn the Gaussmeter setting to batt. Make sure that the voltage is around 0.7 V. (No lower than 0.6 V!)
5. Turn the Gaussmeter setting to 100 mT. Point the probe at the chalkboard, as far away from the solenoids in the class as possible. Record the "offset voltage" which you read on the multimeter. This voltage corresponds to 0 mT.
6. Use the Gaussmeter to determine the magnetic field strength of the solenoid used in Part I of the experiment.a) Turn on your solenoid by running through it the same amount of current you did in part I to get the dot.7. Use the Gaussmeter to determine the magnetic field strength of the Earth. Since it is so small, you will not be able to use your "offset voltage" for reference anymore.
b) Record the voltage you read on the multimeter at 1) the screen of the CRT, and 2) in the middle, 3) 3/4 length, and 4) at the end of the solenoid. In addition, record one more point outside the solenoid.
c) Determine the magnetic field strength of each of these points, and make a graph of B vs length.a) Change the Gaussmeter setting to 10 mT.In addition to your conclusion, answer the questions found in the procedure on page 85:
b) Determine the voltage which is maximum by waving the probe around. Also record the direction where you find this maximum.
c) Find the minimum voltage. Record it and the direction of the minimum.
d) Determine the field strength of the Earth by using the equation |V_max - V_min| / 2.
e) Determine the direction of the field. Does it point to the geographic South (Waikiki)? or North (the North Shore)? Which way did you expect it to go?
They are:
1) Is the approximation that B is constant in the solenoid a good one? Where does this approximation begin to fail?
2) Calculate the magnetic field of the solenoid using B = mu * n * I. Find the % difference between this calculated value and the measured value you got with the multimeter. Why might they be different?
3) Determine the charge to mass ratio using the magnetic field measured with the Gaussmeter. Find the % difference between this value and the q/m you found in part I. Which is better?