Practice Questions on Particles in Magnetic Fields Please answer on file paper. 1. A charged particle of mass m and charge q enters a magnetic field B of flux density 0.45 T as shown: B
a. What is the sign of the charge of the particle?
Use Fleming’s left hand rule – Particle is negative. () Calculate the radius of the path of the particle if the value of the charge of the particle is 1.6 × 10-19 C, the mass of the particle is 1.67 × 10-27 kg and the speed of the particle is 2.5 × 106 m s-1. What do you think the particle is? (3) r = mv ÷ Bq = (1.67 × 10-27 × 2.5 × 106 ) ÷ (0.45 × 1.6 × 10-19) () = 0.058 m (= 5.8 cm) () Particle is a positron. () 2. Blood contains ions in solution. The diagram shows a simplified version of an electromagnetic flow meter to measure the flow of blood through an artery by detecting the movement of ions.
Y Blood flow
The magnet produces a magnetic field strength of 2.0 T, and a potential difference of 600 V is produced between the electrodes X and Y. The cross sectional area of the artery is 1.5 × 10-6 m2 and the separation of X and Y is 1.4 × 10-3 m. The poles of the magnet are of square section; each side has a length of 1.4 × 10-3 m. a. Show that the flux is about 4 × 10-6 Wb A = 1.4 × 10-3 × 1.4 × 10-3 = 1.96 × 10-6 m2 ()
F = 2.0 × 1.96 × 10-6 = 3.9 × 10-6 Wb () b. Draw a diagram to illustrate the idea of flux.
Normal to plane of the coil
c. Show that the force acting on an ion of charge q in the blood which is moving at right angles to the magnetic field is given by F = qE/l. (3) Formulae involved are: F = Bqv, E = Nd/dt, = BA ()
v = length ÷ time = l/t; N = 1 E = B l l/t = Bvl () v = E/Bl F = BqE/Bl () = qE/l d. An ion has a charge of 1.6 × 10-19 C. Show that the force on the ion due to the electric field between X and Y is 6.9 × 10-20 N. (2) F = (1.6 × 10-19 × 600 × 10-6) ÷ 1.4 × 10-3 () = 6.86 × 10-20 N () (QED) e. Why is a potential difference induced? (2) Charged particles moving through a magnetic field () will act like a wire in the magnetic field, inducing a voltage. () f. Calculate the speed of the blood through the artery. v = E/Bl
= 600 × 10-6 ÷ (2.0 × 1.4 × 10-3) () = 0.21 m/s () g. Calculate the volume of blood flowing through the artery every second. V = Al = 1.5 × 10-6 × 0.21 ()
= 3.2 × 10-7 m3 s-1 () 3. A hydrogen ion of mass m and charge q travels at a speed v in a circle of radius r in a magnetic field of flux density B. a. Write down an equation for the speed in terms of these quantities that relates the magnetic forces on the ion to the required centripetal force. (3) F = mv2/r = Bqv () Bq = mv/r () v = Bqr/m () b. Show that the time T for one revolution of the ion is given by the expression: 2m T Bq (2) T = distance ÷ speed = 2r/v () v = Bqr/m T = 2r ÷ (Bqr/m) = 2m/Bq (QED) ()
A cyclotron is a particle accelerator that consists of two D-shaped boxes (called dees). Hydrogen ions are injected into the centre of the machine travel in a spiral path as shown.
Path of charged particles Charged plate P
High frequency ac voltage
The dees are set up in a magnetic field and the machine operates in a vacuum. c. Explain using appropriate physics how the machine works, from when the ions are injected to when they leave the machine at point Q. (6) Marks
There is a clear, balanced and detailed description of how the electron is accelerated in the cyclotron with 11 - 13 points from the examples below. The answer shows almost faultless spelling, punctuation and grammar. It is coherent and in an organised, logical sequence. It contains a range of appropriate or relevant specialist terms used accurately.
There is a description of the interaction between the electron and the magnetic field, with appreciation of the acceleration, with 7 - 10 points described from the examples below. There are some errors in spelling, punctuation and grammar. The answer has some structure and organisation. The use of specialist terms has been attempted, but not always accurately.
There is a brief description of the cyclotron, with 2 - 6 points described from the examples below. The spelling, punctuation and grammar are very weak. The answer is poorly organised with almost no specialist terms and/or their use demonstrating a general lack of understanding of their meaning.
No relevant points
Ion is attracted to electrode with the opposite charge
Ion is accelerated
Path is circular
because the charged particle is in a magnetic field.
Polarity is reversed.
So is attracted by the second electrode.
Force results in acceleration (Newton II)
Since the speed is higher,
The radius is larger.
The path is a spiral because the radius is increasing.
Plate P puts an electrostatic force on the particle
That is equal and opposite to the centripetal force
So the particle leaves Q travelling in a straight line.
d. A particular cyclotron accelerates hydrogen ions in a magnetic field of 0.60 T. i. Calculate the time period for 1 revolution. (2) T = (2 × p × 1.67 × 10-27) ÷ (0.60 ×1.6 × 10-19) () = 1.09 × 10-7 s () ii. Work out the frequency needed to maintain this.
f = 1/T = 1/1.09 × 10-7 = 9.15 × 106 Hz () iii. Why does the fact that the time period is independent of the speed of travel simplify the operation of the cyclotron? (2) The mass and charge of the ion are constant. () So only the magnetic field strength needs to be altered. ()
Total = 35 marks