Magnetic Field and Cathode Ray Tube

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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. 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 velocity making an angle α with the horizontal. Clearly,

 dy  tanα = 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 α = L tan α = 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...
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