Magnetic Fields Tutorial 3 - Force on a Charge



Interaction of Charged Particles

Path of charged particles

Hall Effect



Interaction of Charged Particles with a Magnetic Field

We know that a magnetic field and an electric current interact to produce a force.  Since a current is a flow of charge, it is reasonable to suppose that a magnetic field exerts a force on individual charge carriers.


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.



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:       

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.


If the magnetic field is at an angle q to the magnetic field, the equation is modified to:


F = Bqv sin q


If no angle is mentioned in the question, assume that the angle is 90o.



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



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.




Path of charged particles in a Magnetic Field

We have seen that the force always acts on the wire at 90o, and that gives us the condition for circular motion. 



We can combine the relationship a = v2/r with Newton II to give us:





The v on the left cancels to get rid of the v2 term on the right:




This rearranges to give us:



 Worked Example

An electron passes through a cathode ray tube with a velocity of 3.7 × 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?

Combine F = Bqv and F = mv2/r

to give:


r = 9.11 × 10-31 kg × 3.7 × 107 m s-1 = 0.39 m = 39 cm

       0.47 × 10-3 T × 1.6 × 10-19 C



Question 2

In a particle physics experiment, a detector is placed in a magnetic field of 0.920 T.  A particle is found to produce a circular track of radius 0.500 m.  Other experiments have shown that the particle carries a charge of +1.60 ´ 10-19 C and that its speed was 3.00 ´ 107 m s-1. 


What is the mass of the particle? 

How does it compare to the mass of an electron (9.11 ´ 10-31 kg)? 



The Hall Effect

Magnetism Tutorial 3B is for students for the Welsh Board and Eduqas Syllabuses.  Students studying other syllabuses are, of course, welcome.





The cyclotron is a particle accelerator that relies on this idea.  The machine’s main components are two D-shaped electrodes ("Dees") in an evacuated chamber, placed between the poles of a large electromagnet.


From the top it looks like this:



Notice that the beam of particles is not circular, but a spiral.  This is because the particles are being accelerated by the electric field between each D-shaped electrode (called a dee).  As their speed increases, so does the radius of the curved path.


If a particle of charge q enters one of the dees with a speed v, it will move in a semi-circular path of radius r. 




Ţ rearranging gives us




We can work out from t = s/v what time it takes for the charge to travel:


For 1 revolution:

s = 2pr




The r terms cancel:

Since f = 1/t:


Rearranging gives us:

Question 3

A cyclotron has magnets of flux density 1.50 T and the polarity changes with a frequency of 2.00 MHz.  A large charged particle which has a charge of +2e is inserted into the machine and is accelerated.  Calculate the mass of the particle.



Click HERE to see an animation by Stephen Lucas, a past student of mine.  I am most grateful for his permission to use it.