Question

27B -- Experiment: Magnetic Force on a Moving Charge - qolme Objective: In this module you will observe the trajectory of an electron beam in a uniform magnetic field oriented in various ways to the electron velocity. You will then use your observation of the radius of the electron trajectory to measure the charge to mass ratio for an electron. Equipment: One elm apparatus (Lurge mounted glass bulb with Helmholtz coils); one Leybold or PASCO power supply for electron gun heater and accelerating voltage, one 24V/3A do power supply for magnetic coil current, 8 banana plug leads, one Digital Multimeter. Note that some systems will have an old Leybold power supply for the electron gun and most systems will have a basic HP 24V/3A supply for the current to the coils. We have also added a digital voltmeter to check the high voltage to the gun Theory: This section will lead you through the basic theory for this experiment by Ghe Bowl relating magnetic force to centripetal acceleration for a particle moving at a velocity controlled by an accelerating potential. A schematic of the essential physical elements is shown to the right. This schematic shows electrons being accelerated in a vacuum by a potential difference. An external magnetic field passes through the globe in the Z-direction. Filt 1) First consider the force on a charged particle with charge q and mass m moving at velocity v through a magnetic field - B. a) Write an expression for the force on a charged particle with charge q and mass m moving at velocity through a magnetic field - . Because it is a cross product, this force causes the particle to move in a circle perpendicular to the magnetic field. b) Assume that the magnetic field Ê is in the Z-direction and the velocity is ö = vi+ vy). Find the vector force f. This is the force that causes the circular motion. c) Find the magnitude of the force || from purt (b) above. IL d) In what plane is the force ?

Answer 1

1)

a)The force on a charged particle with charge q in magnetic field is given by:

$\vec{F}=q(\vec{v}\times \vec{B})$

where v is the velocity of particle, and B is the magnetic field.

b) Now, for $\vec{B} = B \hat{k}$ , and $\vec{v} = v_{x} \hat{i}+v_{y} \hat{j}$

$\vec{F} = q\left [ \left (v_{x} \hat{i}+v_{y} \hat{j} \right )\times B\hat{k}}\right ]=q\left [ \left (v_{x}B \left (\hat{i}\times \hat{k} \right )+v_{y}B \left (\hat{j} \times \hat{k} \right ) \right )}\right ]$

$\vec{F} =q\left [ \left (-v_{x}B \hat{j} +v_{y}B \hat{i} \right )}\right ]$

because i x k = -j and j x k = i.

c)

Now, TĒL = VĒL = 57-9vb]2 + (9 45B)? q B J V²+ Vy² - qUB where VE ſva²+ Vy² d) Since the acts in a force and y direction, thus, it any and y is in vý plane 2) The centripetal fore & given by Fc my² r. 3) Now, the magnetic force balances the cortapetal force and accounts for "ootational motion. DO quB = mi r o q V Br m