Experiment 4 Measurement of e by Thomson’s 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? Thomson’s 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. Millikan’s 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 Thomson’s method.

Order nowThis 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 velocity 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 electron’s 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.

For this we require, eE = e v0 B or v0 = E . B The above analysis assumes that the magnetic field B is uniform. However, the magnetic field produced by the bar magnet is non-uniform. Fig. 3 shows the arrangement of the magnets with respect to the CRT. Fig. 3 Note that the CRT is aligned along the magnetic meridian, i. e. S-N direction, which is the direction along which the horizontal component of the Earth’s magnetic field, BE, acts. Since BE and the electron velocities are parallel, there is no deflection produced by the Earth’s magnetic field.

The magnetic field produced by the bar magnets along the path of the electrons will be a function B(x) of the position of the element and will act in the E-W direction as shown in Fig 3. The deflection, y, due to the magnetic force will be in the negative y direction. To calculate the total deflection of the election as it moves from A (anode aperture) to S (screen) we proceed as follows. The radius of curvature r of the electron path y(x) in the presence of the magnetic field is related to the curvature of the path as 1 d2y = . r dx 2 Of course, the radius of curvature will also change with position since the magnetic field changes, i. . , mv0 r = r( x ) = by eq. (9). eB( x ) Thus, d2y eB( x ) = , 2 dx mv0 which upon integrating twice gives, (10) y(x ) = ? ?? ? ? eB( x ) ? dx? dx mv0 ? . The net displacement y at the position of the screen (i. e. x = L0 = L – a) is then ? x eB( x ) ? eI ? ?? mv0 dx? dx = mv0 0 ? 0 ? where we have denoted the double integral of the magnetic field as L0 y= (11) L0 I= ? 0 ?x ? ? ? B( x )dx? dx . ?0 ? (12) When the deflection due to the electric and magnetic field are the same then we can use eqs. (7) and (11) and eliminate the unknown velocity v0 to obtain ?a ? ? + L ? aVy e ? 2 ? = m sI 2 I. 4 Approximate evaluation of I: 13) A calculation of the integral I requires the knowledge of the magnetic field produced by the magnets along the path of the electron in the CRT. We now give an approximate calculation of I assuming that the magnets are very long compared to the length of the CRT (L0). (See Fig. 3). The effect of the distant poles can be ignored. The field at point Q, which will be in a direction normal to the x-axis, is B( x ) = (m + m’ )d (x 2 + d2 ) 3 2 Here, m and m’ are the pole strengths of the N and S poles. The maximum of the magnetic field, Bm occurs at x = 0, Bm = (m + m’ ) d2 Bm d 3 . Thus, we can write, B( x ) =

L0 x (x 2 + d2 ) 3/ 2 (14) The integral I = ? dx ? dx B( x ) can then be evaluated to give 0 0 I = Bm d d 2 + L2 ? d 2 o (15) I. 5 Determination of Bm: The value of Bm is determined from the period of oscillation of the magnetometer needle. The magnetometer is placed at the centre in place of the CRT so that the magnets are at a distance d from it (Fig. 4). The magnetometer needle aligns itself along the resultant 2 2 magnetic field B = BE + Bm where BE is the Earth’s magnetic field acting towards south. Fig. 4 A small disturbance of the needle about the equilibrium position causes it to oscillate.

The angular frequency of small oscillations can be easily shown to be (Exercise 7) µB , where µ is the magnetic moment of the needle and I its moment of inertia, ? = I I . Now, the resultant and hence the time period of small oscillations T = 2? µB 2 2 magnetic field B= BE + Bm = Bm / sin ? 0 . Thus Bm = 4? 2 I sin? o T2 µ In the absence of the magnets B = BE (due to the Earth’s magnetic field) and the time I 4? 2 I period T0 = 2? giving = T02 BE . µBE µ Thus, Bm = T02 BE sin ? 0 . T2 (16) With a determine of Bm the value of the I (eq . 15) can be calculated and hence from eq. 13) the value of e/m. II 1. Set-up and Procedure : PART A Place the magnetometer on the wooden box enclosing the CRT. Rotate the dial so that 00– 0 0 position is perpendicular to the length of the CRT. Next rotate the CRT with the magnetometer on it so that the magnetometer needle aligns along 0 0– 00 position. In this position the CRT is aligned along the magnetic meridian (N-S position) while the scales attached perpendicular to the CRT (for magnet mounting), are in E-W position. Switch on the power supply and adjust the intensity and focus controls to obtain a fine spot on the CRT screen. Note: The position of the spot may not be at the centre of the CRT screen. Note down the initial position of the spot). Choose a value of the deflection voltage and note down the deflection of the spot. (Note: The deflection has to be taken with respect to the initial spot position). 2. 3. 4. 5. 6. 7. Now extend the scales attached to the CRT and place identical bar magnets (as shown in Fig. 3) and move the magnets symmetrically along the length of the scales until the spot deflection becomes zero (i. e. , the spot returns to its initial position). Note the value d of the distance of the magnet poles from the centre of the CRT.

Reverse 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 electron’s 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 Earth’s 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. “Advanced Practical Physics for Students”, B. L. Worsnop and H. T. Flint, Metheun London, 1942. “Physics”, M. Alonso and E. J. Finn, Addison Wesley, 1992. 8. 9. 10. References: Experiment 4: Measurement of e by Thomson’s 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 Earth’s 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