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Magnetic force I. Objectives. In this lab you will explore the force exerted by a magnetic field on a beam of electrons. II. Equipment. Electron beam tube, high voltage power supply (model TEL 813), Helmholtz coils, Helmholtz coils 10 Volts power supply, Ammeter, magnetic compass. III. Introduction. IIIa. Motion of a charge in a uniform magnetic field. Consider an electron which is injected in a region with a uniform magnetic field B , with its velocity at right angles to the magnetic field, as shown in Fig.1. The laws of mechanics predict that the electron will move on a circular path of radius R which is given by the equation: R mv Be (eqs.1) Here v is the speed of the electron, m is the electron mass ( m 9.111031 kg), and e is the electron charge ( e 1.6 1019 C). The magnetic force F evB is also shown in Fig.1. Note that this force points towards the center C of the circular orbit, something which is counterintuitive. We note that we have a similar situation with the moon. The earth pulls the moon towards it but the moon does not fall towards the earth. Instead it orbits around the earth. The speed v of the electron remains constant. IIIb. Generation of an electron beam; the electron beam tube also known as “cathode ray tube” or CRT for short. The tube used to generate a beam of electrons all of which move to the right with velocity v is shown in Fig.2. It is a sealed glass tube from which the air has been evacuated and which contains two metal electrodes called the “cathode” and the “anode”. The cathode is a filament which is heated by applying a 6.3 AC voltage because a hot metal emits electrons in the surrounding vacuum. If we apply a voltage difference V between the anode and the cathode making sure that the anode voltage is higher than that of the cathode, the electrons are accelerated to a velocity v by the time they reach the anode. The electron velocity v at the anode is given by the following equation: v 2eV m (eqs.2) Here, m the electron mass (give the value) and V is the potential difference between the anode and the cathode. The anode has an opening at its center so that the electrons can pass through it and travel to the right past the anode with constant velocity v . This part of the tube has the form of a large bulb. In this region we generate a constant magnetic field B by flanking the bulb on both sides with two Helmholtz coils. The operation of the Helmholtz coils with be given in section IIIc below. We can determine B by 1 measuring the current i that passes through the coils. If we combine eqs.1 with eqs.2 and eliminate v we get: B 1 2V R e/m (eqs.3) 1 , all the points R fall on a straight line that passes through the origin and has a slope S given by the equation: If we measure the orbit radius R for each value of B and plot B versus S 2V e/m (eqs.4) Past the anode we have an xy-grid made of mica which is coated with a fluorescent material that glows blue when the electron beam hits it. This makes it possible for us to see the electron beam. The smallest division on the grid is equal to 0.2 cm. The origin O of the grid is located at the center of the anode as shown in Fig.3. If the bulb were larger we would be able to see the full circular orbit of the electron. In this particular tube we can see only a section of the circular orbit between point O and point P whose coordinates x and y we can measure on the grid. Using simple geometry arguments we can show that: R x2 y 2 2y (eqs.5) IIIc. Helmholtz coils. These are shown in Fig.4. Each has a diameter D and consists of N turns of wire. They are positioned so that they have a common axis and the distance between the coil centers is adjusted so that it is equal to half the diameter D . Under these conditions the Helmholtz coils generate a magnetic field along the common axis of the coils which is uniform in the vicinity of the midpoint between the coil centers. The magnetic field B is given by the equation: B 160 Ni D 125 (eqs.6) Here the current i is measured in Amps, B is measured in tesla, D 0.136 m, N 320 , and 0 is a constant equal to 1.256 106 T.m/A If we substitute the numerical values for 0 , D , and N we get the equation: B in Tesla 4.23 106 i in mA (eqs.7) 2 IV. Procedure Use the circuit shown in Fig.2. Align the electron tube along the north-south direction. a. Use an anode-cathode voltage of 2000 Volts. With zero current through the Helmholtz coils you should be able to see a blue streak at the center of the grid. The blue light is generated when the electrons collide with the fluorescent material on the grid. We will measure at x = 10 cm i.e. at the far right end of the grid. b. Pass some current i through the Helmholtz coils so that the electron beam is deflected upwards ( y 0 ) by the magnetic force. If the electron beam moves downwards instead change the direction of the current. Adjust the current i in the Helmholtz coils so that the y-coordinate at x = 10 cm is equal to 0.2 cm. Record the current i in the appropriate column of your data table. c. Reverse the direction of the current in the Helmholtz coils so that the electron beam is deflected downwards ( y 0 ) . Adjust the current i in the Helmholtz coils so that the ycoordinate at x = 10 cm is equal to -0.2 cm. Record the current i in the appropriate column of the data table. d. Repeat steps IVa and IVb for y = 0.4, 0.6, 0.8, 1.0, 1.2, 1.4, 1.6, 1.8, 2.0 , 2.2, and 2.4 cm. V. For the report i i and enter the value in 2 the corresponding column of the data table. Using equation 7 Calculate the magnetic field B that corresponds to iav and enter the value in the value in the corresponding column of the data table. a. For each value of y calculate the average current iav b. Plot B versus 1 . The points should lie on a straight line that passes through the R origin. c. Use the least squares fit method to determine the experimental value Sexp of the slope of the graph. d. Use equation 4 to calculate the theoretical value Scalc of the slope of the graph. e. Find the percentage difference Sexp Scalc Scalc calculated values of the slope. 3 100 between the experimental and the B R v C F electron Fig.1 Fig.2 cathode 6.3 V grid anode electron path e V High voltage power supply 4 Fig.3 B C R P (x,y) Anode Cathode v O v Figure 4 5