Magnetic Field and Cathode Ray Tube Essay Example For Students

Magnetic Field and Cathode Ray Tube Essay Experiment 4 Measurement of e by Thomsons bar magnet method. m Apparatus: Cathode ray tube (CRT) with power supply unit, one pair of bar magnets, high resistance voltmeter, magnetometer, stopwatch. Purpose of experiment: ? e? To measure the specific charge, i. e. charge to mass ratio ? ? , of an electron using ? m? Thomsons bar magnet method. Basic methodology: Electrons in a CRT are deflected in the vertical direction by applying a potential between the vertical deflection plates of the CRT. A magnetic field perpendicular to the deflecting electric field is produced using a pair of bar magnets. The position of the magnets is adjusted so as to cancel the deflection of the electrons. The knowledge of the deflecting potential and the magnet field of the bar magnets leads to a calculation of the specific charge. I. Introduction We have learnt that the electron has a negative charge whose magnitude e equals 1. 6 ? 10? 19 Coulomb and mass (m) equal to 9. 1 ? 10 ? 31 Kg. Millikans Oil Drop method enables us to measure the electron charge but the mass of the electron can not be measured directly. It is calculated by measuring the value of e/m. The aim of this experiment is to determine value of e/m by Thomsons method. This involves the motion of an electron in a cathode ray tube (CRT). A simplified form of a cathode ray tube is shown in Fig. 1. The electrons are emitted from the cathode and accelerated towards the anode by an electric field. A hole in the accelerating anode allows the electrons to pass out of the electron gun and between the two sets of deflection plates. The metallic coating inside the tube shields the right end free of external electric fields and conducts away the electrons after they strike the fluorescent screen where they form a luminous spot. Plates for Plates for vertical Focusing Anode horizontal deflection Accelerating deflection Anode Control Grid Electron Beam Metallic Coating Fluorescent Screen Heater Cathode Electron Gun Fig. 1 I. 2 This experiments can be divided into the two followings parts: 1. 2. The electric field (E) is applied alone. This produces a deflection of the electron beam. A magnetic is simultaneously applied along with the electric field so the deflection produced by the electric field is exactly counter-balanced by that produced by the magnetic field. As a result the spot made by the electron on the fluorescent screen returns back to the central position. Fig. 2 Fig. 2 Let us consider an electron moving in the direction of magnetic meridian (say x. -axis) with a velocity v0 m/s after passing through the accelerating anode. Under the action of the electrostatic field E = V/s (s being the vertical distance between the plates VV and V the deflecting voltage) each electron, as it passes between the plates, is acted upon by a perpendicular force eE. As a result the electron moves along a parabolic path AB (Fig. 2). The equation of motion is ? d2y? m? 2 ? = eE , ? dt ? ? ? which, upon integrating once with respect to time, gives (1) dy ? v0 ? ? = (eE / m )t ? dx ? (2) where v0 = dx/dt is the constant horizontal velocity. Here we also used the initial condition dy/dx = 0 at point A and time t=0. At any point distant x (=v0t) from point A in the field between the plates VV, eq. (2) gives dy ? eE ? ?x =? dx ? mv0 2 ? ? ? (3) On leaving the electrostatic field at point B (i. e. x=a), the electron moves along the tangential path BC with its velo city making an angle ? with the horizontal. Clearly, ? dy ? tan? = tan FBC = ? ? = ? dx ? at po int B ? eE ? ? ? ? mv 2 ? a ? 0 ? = tangent to the curve AB at point B (4) The electron finally strikes the screen at the point C (Fig. 2). The total vertical deflection of the electron y = CF + FO. Now CF = BF tan ? = L tan ? = eEaL . 2 mv0 (5) On the other hand, by eq. (3), we have ? eEx ? eEa 2 ? dx = BD = ? dy = ? ? ? mv 2 ? 2 mv02 ? 0 ? (6) Therefore the total displacement (y) in the spot position on the screen S due the application of electric field between the plates VV is y = CF + F O = CF +BD y= Thus eEa ? a ? + L? 2 ? mv0 ? 2 ? (7) e = m 2 v0 y ? a ? Ea? + L ? 2 ? ? (8) Hence, if the velocity of electron along the X-axis (v0) is known, the value of e/m can be calculated. I. 3 Let B be the magnetic field produced by the two bar magnets placed symmetrically on either side of the cathode ray tube at a distance d from it. The magnetic field of the bar magnets will be in the east-west direction. The magnetic force on the electron is given by F = ? e v ? B . This sets up a force Bev0 on the moving electron along the y direction. As a result, the electrons path becomes circular, with radius of curvature r given by 2 mv0 = Bev0 . r ( ) (9) When the forces on the electron beam due to electric and magnetic field are equal and opposite, the electron beam will be undeflected. Discrimination Everywhere! EssayReverse the deflection voltage and (with magnets removed) note the deflection of the spot. Place the magnets on the scale and find the value of d for which the spot returns to its initial position. Repeat the above steps for three different spot positions (Note: The deflection voltage should not exceed 375 Volts) PART B: Determination of time period of oscillation of magnetometer needle Align the wooden arm on which the magnetometer is placed along the magnetic meridian and place the magnets along the scales in the EW direction at the same distance d as in Part A. Note the equilibrium deflection ? 0. With a third magnet, slightly disturb the needle from its equilibrium position and measure the time period of oscillations T. Now remove the magnets and let the needle come to equilibrium at 00 – 0 0 position. Disturb the needle about this equilibrium position and measure the time period T0 of the oscillations. 1. 2. 3. 4. 5. 1. 2. 3. 4. Precautions: The cathode ray tube should be accurately placed with its longitudinal axis in the magnetic meridian. The spot on the screen should not be allowed to remain at a given position on the screen for a long time. There should not be any other disturbing magnetic field near the apparatus. While taking the observations for time periods, the maximum angular displacement of the magnetic needle should not exceed 40-50 degrees. Exercises and Viva Questions. Study the working of a CRT. What is the a typical value of accelerating voltage used in a CRT ? Estimate the velocity v0 of the electron. What will happen to the spot if a sounsidally time varying voltage is applied to the deflecting plates VV or HH? What will happen if such a voltage (of the same frequency) is applied simultaneously to the horizontal and vertical deflection plates? Draw a neat diagram showing the 3-dimensional orientations of vectors of the electrons horizontal velocity, the electric field, the magnetic field, the electric force on the electron and magnetic force as the electron moves in the CRT. Orient your diagram according to the experimental set-up. If the deflecting voltage is switched off but the bar magnets kept in place, will there be a deflection of the spot? Describe qualitatively the motion of the electron in the CRT from aperture to screen. III. 1. 2. 3. 4. 5. 6. 7. Describe the motion of the electron in the CRT in the presence of the deflecting voltage magnetic fields of the magnets. What is the effect of the Earths magnetic field on the electron motion? What would happen if the apparatus were rotated by 90 ° so that the CRT is along the E-W direction. Consider a dipole  µ aligned with a magnetic field. If the dipole is given a small angular displacement, then it experiences a restoring torque ? =  µBsin? where ? is the angle between the dipole and magnetic field. Considering small displacements ? , show that  µB the dipole will oscillate about the equilibrium with angular frequency , I being its I moment of inertia. Recalculate the integral I (eq. 12)) assuming that the magnetic field of the magnet is a constant B = Bm. Use this to recalculate the specific charge e/m. Does our approximate evaluation of I improve the evaluation of e/m? What are the sources of error in this experiment? How does your result compare with the e/m measurement by Thomson’s method? Which experiment is more accurate? 1. 2. â€Å"Advanced Practical Physics for Studentsâ₠¬ , B. L. Worsnop and H. T. Flint, Metheun London, 1942. â€Å"Physics†, M. Alonso and E. J. Finn, Addison Wesley, 1992. 8. 9. 10. References: Experiment 4: Measurement of e by Thomsons bar magnet method m Observations and Results Observations: Constant values: Length of plate, a = 2 cm Distance to screen from plate, L = 16. 0 cm Distance between the plates, S = 0. 4 cm Horizontal component of Earths magnetic fields BE = 3. 53 ? 10? 5 T. PART A: Measurement of deflection y: Initial position of spot, y0 = _____________ cm (specify +ve or –ve). Table 1 Displaced position of spot y1 (cm) Direct Reverse Displacement of spot y = y 1? y0 Direct Reverse Position of magnet d (cm) Direct Reverse Applied Voltage V (volts) Mean displacement y (cm) Mean d (cm) PART B: Determination of Time period Table 2 No. of oscillations = _______________ With Magnets Without Magnets S. No Total time T0 d =_____cm d =_____cm d =_____cm ?0 = ____deg Total time T ?0 = ____deg Total time T ?0 = ____deg Total Time T 1 2 3 Mean T0 Mean T All time measurements are in seconds Calculations  §a  · ? + L ? a Vy e  ©2 ? = m SI 2 Displacement d(cm) Bm = T BE sin ? 0 T 2 0 2 2 I = Bm ? d d 2 + Lo ? d 2 ?  «  »  ¬ ? Results: Calculated value of specific charge of electron Standard value of e = _____________ C/kg. 1. 759X1011 m e = ___________ C/kg. m % error in e = ___________________ m