7 Motion of a charged particle in the magnetic field

1

To investigate the effect of a magnetic field on moving charged particles;

2

Why the electron beam is deflected as it passes through charged metal plates?

An electron beam tube (Figure) can be used to demonstrate the magnetic force on a moving charge. A beam of electrons is produced by an ‘electron gun’, and magnets or electromagnets are used to apply a magnetic field.

The electron gun shoots out a beam of electrons across an evacuated tube. It hits a fluorescent screen placed in its path and when it does the screen glows. If there is no voltage between the two plates the beam will go along the middle of the scale.
Beams of electrons (called cathode rays) move in straight lines in a vacuum when there is no electric or magnetic field.

If a voltage is applied across the plates with the bottom plate positive, the electron beam will be attracted downwards towards the positive plate showing that the beam really is made up of negative particles. If the top plate is made positive the beam will be attracted upwards.

A bar magnet can now be held at the side of the tube and you will see that the beam of electrons is deflected up or down depending which way round you hold the magnet. The same thing will happen of course if you use an electromagnet

The path of the electrons in a magnetic field is a circle.

The factors that determine the size of the force on a moving charge in a uniform magnetic field. It will depend on:
■■ the magnetic flux density B (strength of the
magnetic field)
■■ the charge Q on the particle
■■ the speed v of the particle.

The magnetic force F on a moving particle at right angles to a magnetic field is given by the equation:
F = Bqv
The direction of the force can be determined from Fleming’s left-hand rule. The force F is always at 90° to the velocity of the particle. Consequently, the path described by the particle will be an arc of a circle.

Suppose we have such a particle with a charge q, moving at a speed v, at right angles to a magnetic field of flux density B. In a time t, the charge will move a distance L = v∙t and is equivalent to a current I = q / t.
Force on the current
F = BIL = B ∙ q / t ∙ v ∙ t = Bqv
If the field and current are at an angle q, then the formula will be modified to F = Bqv sinϑ

The force F is always at right angles to the particle’s velocity v, and its direction can be found using the Fleming’s left-hand rule.

I Round – Multiple choice Question [2]
II Round – Test Yourself [6]
III Round – Problem solving [3]

A charged particle is situated in a region of space and it experiences a force only when it is in motion. Which of the following states the field or fields enclosed in the region correctly?
A) An electric field only
B) Both a magnetic field and an electric field
C) Both an electric field and a gravitational field
D) Both a magnetic field and a gravitational field
E) A magnetic field only

E [1]
Explanation: Magnetic fields exert forces on charged particles only if those particles are moving. [1]

2. What is the direction of the velocity of a negative charge that experiences the magnetic force shown in each of the three cases in, assuming it moves perpendicular to B?

(a) East (right) [1]
(b) Into page [1]
(c) South (down) [1]

3. What is the direction of the magnetic field that produces the magnetic force on a positive charge as shown in each of the three cases in the figure below, assuming  is perpendicular to ?

(a) Into page [1]
(b) West (left) [1]
(c) Out of page [1]

4. A cosmic ray proton moving toward the Earth at  5.0 ∙ 107 m/s experiences a magnetic force of  1.70 ∙ 10-16 N. What is the strength of the magnetic field if there is a  450  angle between it and the proton’s velocity?

B = F / qv [1]
B = 1.70 ∙ 10-16 N / 1.6 ∙ 10-19 C × 5.0 ∙ 107 m/s [1]
B = 3.0 ∙ 10-5 T [1]

What has been learned
What remained unclear
What is necessary to work on