Tutorial 5 A - Magnetic Forces on  Electrons in a Vacuum

Learning Objectives

Electron beams make circular paths in magnetic fields

The force on an electron from a magnetic field can be counteracted by that of an electric field.

Key Questions

What is the path of an electron in a magnetic field?

How do we measure the speed of an electron?

Can magnetic fields interact with electric fields?

Magnetic Force on an electric charge

We have seen how a magnetic field will produce a force on a current in a wire.  A current is defined as flow of unit charge in unit time.  In other words, the charges are moving.  If the charges are not moving, then there is no force.


We can take away the wire, and have the electrons jumping a gap between a negative and positive terminal in a vacuum.  If we apply a magnetic field, they will experience a force.  We can do the same with positive ions and they too will experience a force in a magnetic field.  A vacuum is essential because the electrons will collide with molecules and rapidly lose their energy.


We find that in a magnetic field, the force acts on a stream of electrons always at 90o to the direction of the movement. Therefore the path is circular.  The direction is determined by Fleming's Left Hand Rule.


Remember that the direction of the electrons’ movement is in the opposite direction to the conventional current.  So if the electrons are going from left to right, the conventional current is going from right to left.


When using Fleming’s Left Hand Rule, the current is conventional.


Consider a charge q moving through a magnetic field B at a constant velocity v.  The charge forms a current that moves a certain distance, l, in a time t.

We know:       

Velocity = distance ÷ time

Current = charge ÷ time

F = BIl

So we can substitute into this relationship to give us:

            F = B × (q/t) × vt = Bqv

So the formula now becomes:

 F = Bqv

[ F- force in N; B – field strength in T; q – charge in C; v ­– speed in m s-1]

The charge is usually the electronic charge, 1.6 × 10-19 C.


Question 1

An electron accelerated to 6.0 ´ 106 m s-1 is deflected by a magnetic field of strength 0.82 T that is acting at 90o.  What is the force acting on the electron?  Would the force be any different for a proton?



If the stream of charges is at an angle other than 90o to the magnetic field, the formula becomes:

If the charged particle enters at an angle of 90o to the field, the path is circular.  If the angle is not 90o, the path is helical.

Question 2

An electron accelerated to 6.0 ´ 106 m/s is deflected by a magnetic field of strength 0.82 T.   It makes an angle of 76o to the field.

(a) Calculate the force acting on the electron now.

(b) Calculate the forward velocity as it makes its helical path along the magnetic field.



We have seen that the force always acts on the particle at 90o, and that gives us the condition for circular motion.  The force F is the centripetal force.
We can combine the relationship for centripetal acceleration:

 with Newton II (F = ma) to give us:


F = Bqv     

We can equate these two equations to give:

We can do some cancelling to give:


This rearranges to give us:

Worked example

An electron passes through a cathode ray tube with a velocity of 3 × 107 m s-1.  It enters a magnetic field of flux density 0.47 mT at a right angle.  What is the radius of curvature of the path in the magnetic field?



Substitute the numbers:



Question 3

In a particle physics experiment, a detector is placed in a magnetic field of 0.92 T.  A particle is found to produce a curved track of radius 0.5 m.  Other experiments have shown that the particle carries a charge of +1.6 ´ 10-19 C and that its speed was 3.0 ´ 107 m/s.  What is the mass of the particle?  How does it compare to the mass of an electron (9.11 ´ 10-31 kg)?    



How do we measure the speed of an electron?

If we go back to what we know about electrical quantities, we know that energy (J) = charge (C) × voltage (V).


E = qV


We also know from basic mechanics that energy of a moving object is its kinetic energy, which is given by the equation:

 The energy in a moving electron is always kinetic, so it doesn't take a genius to see that:

We can rearrange the equation to give us the speed:

Or, if you prefer:

In other words, we simply measure the voltage.


Question 4

An electron is accelerated by a voltage of 1300 V.  What is its speed?

Give your answer to an appropriate number of significant figures.

Mass of an electron = 9.11 × 10-31 kg; Charge on an electron = 1.602 × 10-19 C



Then we can use the equation below to give us the expected radius of curvature:


If we chose to be very clever, we could equate the two relationships, but there is really no point.  You would end up with a lot of square terms and you would forget to square one of them.  Keep it simple.


Question 5

What is the radius of curvature of the electron above, if the magnetic field has a flux density of 0.035 T?



How do we know the magnetic field?

In all the problems so far, we have stated what the magnetic field should be.  But magnetic field is not easy to measure directly.  It can be done with a Hall probe, which we will look at later in this tutorial.  However the experiment that give rise to what we have looked at in this tutorial is conducted using a magnetic field generated using Helmholtz Coils.



These are arranged around the electron acceleration tube like this:


Image by courtesy of Ansgar Hellwig, Wikipedia.

The magnetic field is uniform as shown in the picture:

© 2010 Geek3 / GNU-FDL, commons.wikimedia.org/wiki/File:VFPt_cylindrical_magnet_thumb.svg


The magnetic flux density is rated to the current in a pair Helmholtz coils is given by this equation:


Maths note

A number raised to the power 3/2 is the square root of that number that has been cubed.  For example:

43/2 = Ö(43) = Ö(64) = 8


We can of course write this as 41.5 = 8, which we can put into any scientific calculator.


A useful little dodge here is that (4/5)3/2 = (0.8)1.5 = 0.71554


The other terms in the equation are:

B = magnetic flux density (T);

m0 = permeability of free space = 4p × 10-7 H m-1;

N = number of turns;

R = radius of the coil (m).


The coils are set apart at a distance that is the same as the radius.  This  formula enables us to calculate the magnetic flux density when we measure the current, easy enough with an ammeter.


Question 6

A pair of Helmholtz coils has a total of 500 turns and is carrying a current of 2.5 A.  The radius is 7.5 cm and the two coils are 7.5 cm apart.  Calculate the magnetic flux density.



Keep the calculations in simple steps when you try the next example.

Question 7

In a demonstration electrons are accelerated by a voltage of 1200 V into a uniform magnetic field that is generated by two Helmholtz coils of 500 turns and radius 7.5 cm (and 7.5 cm apart).  The radius of the electron path is measured at 3.4 cm. 

(a) Show that the speed of the electrons is about 2 × 107 m s-1;

(b) Calculate the magnetic flux density of the field;

(c) Calculate the current in the Helmholtz coils.


Mass of an electron = 9.11 × 10-31 kg; Charge on an electron = 1.6 × 10-19 C

m0 = 4p × 10-7 H m-1



Electron paths in other types of force field

Electrons have a small but definite mass (9.11 × 10-31 kg).  Like all objects that have mass, they are attracted by gravity fields.  A stationary electron, if released in a vacuum, will fall vertically downwards and accelerate at 9.81 m s-2.  If it is horizontally travelling at a constant speed, its path will be parabolic, just like any other projectile.



We can apply the equations of motion to study its movement.  You will have done those in mechanics.  If you need to revise projectile motion, click on the link to my A-level website below:




In reality the speed of an electron is so fast that the deviation from a straight line is minimal in a vacuum tube in a laboratory.  The calculation below shows this.


Question 8

An electron is accelerated by a voltage of 1.5 V.  It travels along a horizontal path that is 15 cm long. 

(a)  Show that the speed of the electron is about 730 km s-1

(b)  Calculate how far the electron deviates from the horizontal path.


Mass of an electron = 9.11 × 10-31 kg; Charge on an electron = 1.6 × 10-19 C



The path of an electron in an electric field is also parabolic.  Here we see the electric field between two plates:


The electron is negative and is attractive to the positive plate.  For more about motion of electrons in electric fields, see:


The electric field strength is defined as force per unit charge.  So electric field strength, given the Physics code E has the formula:

For a uniform field, electric field is the potential difference per unit length.

Unlike gravity, we can control the electric field strength simply by turning up the voltage.



Be careful - the accelerating voltage may be different to the electric field voltage.


Once the electrons have left the electric field, they move in a straight line.


We use the same problem solving strategy as we do with projectiles.  If we know the electric field strength, we can work out the force easily by equating the electric field equations:



When we know the force, we can apply Newton II to give us the acceleration.  This is what we will do in the next calculation.


Question 9

Electrons are accelerated by a potential difference of 2000 V.  They pass from left to right on the centreline between two horizontal plates that are separated by a distance of 3.0 cm  The plates are 5.0 cm long and the top plate has a potential of 1000 V, while the bottom plate has a potential of 0 V. 


(a) the speed of the electrons;

(b) the force acting on each electron;

(c) the acceleration of each electron;

(d) the resultant velocity of the electrons in the beam;

(e) the direction relative to the horizontal.

(f) By considering the displacement of the beam, comment on whether the values you have worked out would actually be observed.


Mass of an electron = 9.11 × 10-31 kg; Charge on an electron = 1.6 × 10-19 C




Balancing Electric and Magnetic Fields

This experiment is done using apparatus like this:

If we apply a magnetic field, this is what we will see on the fluorescent screen.

The path is circular


Question 10

Explain why the circular path is below the electron beam.  What would happen to the circle if the magnetic flux density were to be increased?



If we had an electric field, the path would be parabolic.

The electrons are attracted to the positive plate.


What happens when both an electric field and a magnetic field are applied?  At a certain point, the effect of the magnetic field is counteracted by the effect of the electric field.  The result of this is that the overall force on the electrons is zero.  So they are not deviated from their original path.


Question 11

What do you think would happen if the magnetic field were coming out of the screen towards you instead of going away from you?



We can therefore say that:

Force from electric field (N) + Force from magnetic field (N) = 0


Force from electric field = force from magnetic field

Since the definition of the electric field is:

So we can say:

F = Eq


We also know that:

F = Bqv


So we can equate the two equations above by writing:


Eq = Bqv


We can work out the speed v by considering the kinetic energy of each electron.  The kinetic energy of each electron is given by:


Ek = qV1


Voltage V1 is the accelerating voltage from the electron gun.  It is NOT necessarily the same as the voltage across the plates, V2.    We also know that the kinetic energy of the electrons is given by:

Rearranging we get:

To substitute the v term into the equation Eq = Bqv, we need to square the equation.  The q terms cancel out, so:


Now we can substitute:

There is an important quantity to physicists here, the charge to mass ratio, or the specific charge, q/m.  If you have done A level Physics, you will have met this quantity before in Particle Physics.  We can rearrange the formula to give us q/m:

We also know that E = V2/d for the uniform electric field, so we can change this equation to:


To get a value for q/m, we keep V1 constant, and vary V2 and B.  The equation is:


This will give us a straight line:

Question 12

The experimental data to give this graph are given below:

V1 = 300 V

d = 0.025 m


Use the graph above to calculate the gradient.  Hence work out a value for q/m.  Give the units.  Give your answer to an appropriate number of significant figures.



We can work out the magnetic flux density from the current using the Helmholtz coil formula:


This experiment was carried out by J J Thomson in 1897.  It form one of the key pieces of evidence for the charge on an electron.  The charge on an electron was worked out by R A Millikan in 1903.  Then the mass of the electron can be worked out.  Both of these experiments are based on straight-forward physics principles, but the execution of the experiments are fiddly and tedious.


Magnetic coils are used to focus beams of electrons in the cathode ray oscilloscope and the electron microscope.




Tutorial 5B