1. Book Reference : Pages 113-115 1.To understand that the path of a charged particle in a magnetic field is circular 2.To equate the force due to the magnetic field to the centripetal force 3.To examine practical applications of circular displacement of particles

2. In the previous lesson we have seen that a moving charged particle is deflected in a magnetic field in accordance with Fleming’s left hand rule Electron Gun Magnetic field coming out of the page Initial Path of the Electron Force Conventional Current Electron Gun Force Conventional Current Electron Gun Force The force acts perpendicular to the velocity causing the path to change.... The force acts perpendicular to the velocity causing the path to change.... The force acts perpendicular to the velocity causing the path to change.... Circular motion is achieved Conventional Current

3. The force on the moving charged particle is always at right angles to the current velocity... Circular motion! As always with circular motion problems we are looking for a force to “equate” to the centripetal force From the last lesson we saw that the force on a charged particle is F= BQv BQv = mv 2 / r  r = mv/BQ The path becomes more curved (r reduced) if the flux density increases, the velocity is decreased or if particles with a larger specific charge (Q/m) are used Velocity BQv

4. A beam of electrons with a velocity of 3.2x10 7 ms -1 is fired into a uniform magnetic field which has a flux density of 8.5mT. The initial velocity is perpendicular to the field. Explain why the electrons move in a circular orbit Calculate the radius of the orbit What must the flux density be adjusted to if the radius of the orbit is desired to be 65mm [21mm, 2.8mT]

5. The magic A2 crib sheet quotes the following : Charge on an electron (e) = -1.6 x 10 -19 C Electron rest mass (m e )= 9.11 x 10 -31 kg However it also quotes... Electron Charge/Mass ratio(e/m e ) = 1.76x10 11 Ckg -1 Two things.... Do not be phased by this since they could quote this for another particle... Simply inverting it (m/Q) for this sort of calculation saves one step in the calculator!

6. Last lesson we saw the CRT (Cathode Ray Tube) Such devices can also be called Electron guns Thermionic devices The cathode is a heated filament with a negative potential which emits electrons, a nearby positive anode attracts these electrons which pass through a hole in the anode to form a beam. This is called Thermionic emission. The potential difference between the anode and cathode controls the speed of the electrons.

7. These are machines which can be used to analyse the types of atoms, (and isotopes) present in a sample r = mv/BQ The key to how it works is the effect that the mass has on the radius of the circular motion while keeping the velocity and flux density constant

8. First we need to ionise the atoms in the sample so that they become charged... Electrons are removed yielding a positive ion

9. A component known as a “velocity selector” is key to obtaining a constant velocity The +ve ions are acted upon by both an electric field & a magnetic field. Only when they are equal & opposite do the ions pass through the slit Collimator with slit The electric field is given by F = QV/d & the magnetic field given by F = BQv Only ions with a particular velocity will allow QV/d = BQv & hence only ions with that particular velocity make it through the slit. Note that it is also independent of charge since the Qs cancel Positive Ion

10. Cyclotrons are a method of producing high energy beams used for nuclear physics & radiation therapy An alternating electric field is used to accelerate the particles while a magnetic field causes the particles to move in a circle, (actually a spiral since the velocity is increasing) Compared to a linear accelerator, this arrangement allows a greater amount of acceleration in a more compact space

11. Two hollow D shaped electrodes exist in a vacuum. A uniform magnetic field is applied perpendicular to the plane of the “Dees” Charged particles are injected into a D, the magnetic field sets the particle on a circular path causing it to emerge from the other side of this D & to enter the next. As the particle crosses the gap between the Ds the supplied current changes direction (high frequency AC) & the particle is accelerated, (causing a larger radius)

12. Assuming Newtonian rather than relativistic velocities apply... The particle leaves the cyclotron when the velocity causes the path radius to equal the radius R of the D v = BQR/m The period for one cycle of the AC must approximate to the time for one complete circle (2  R) using s=d/t for t T = 2  Rm/BQR  The frequency of the AC will be f=1/T f=BQ/2  m