• To identify that the force on a charged particle travelling perpendicular to a uniform magnetic field can be determined using the equation F = Bqv
  • To understand and appreciate that charged particles moving in a uniform magnetic field undergo circular motion

With the knowledge that a current carrying conductor placed perpendicular to, and in, a magnetic field experiences a force, the appreciation of what may happen to a charged particle can begin to be analysed and understood.

Current after all is the rate of flow of charged particles and is denoted by the equation:

I = \frac{\Delta q}{\Delta t}

It is due to the motion of these charged particles within the field that causes the force. We should therefore be able to derive a new equation for the force that single charged particle experiences when in a magnetic field.

Firstly, we know from the previous page  that if there is no current then there is no force. This then implies that either:

  1. When no charge is present, there will be no force; F = 0 N when q = 0 C
  2. When the charges are stationary, have no velocity, there will be no force; F = 0 N when v = 0 ms^{-1}

An equation linking force to charge, magnetic flux density and velocity should therefore exist.

F = BIl       (1)

I = \frac{\Delta q}{\Delta t}     (2)

Substituting (2) into (1) gives:

F = B \frac{\Delta q}{\Delta t}l

Rearranging this slightly for a single constant charged particle gives us:

F = Bq \frac{l}{\Delta t}

Since velocity is the distance per unit time v = \frac{l}{\Delta t} we can write:

F = Bqv

F is the force, measured in Newtons, N
B is the magnetic flux density, measure in Tesla, T
q is the charge, measured in Coulombs, C
v is the velocity of the charged particle measured in metres per second, ms^{-1}

There we have it, an equation linking force to charge, magnetic flux density and velocity does exist. This equation fits the conditions stated earlier. When there is no charge, F = B \times 0 \times v = 0 N , similarly, when there is no velocity, F = B \times q \times 0 = 0 N . Similarly, if there is no magnetic flux density then there is nothing causing the force in the first place; F = 0 \times q \times v = 0 N .

This force is the same as the one in Fleming’s Left Hand rule. It is exerted on the charge perpendicular to the magnetic field and perpendicular to the direction of the velocity (In Fleming’s Left Hand rule it is current though).

Fleming’s Left Hand rule:

This links the second finger to conventional current which is the flow of ‘positive charge’. If singular charged particles flow in a magnetic field instead, Fleming’s Left Hand rule becomes:

The following diagram helps put this force into perspective.

A positive charged particle moving with a constant velocity v into a uniform magnetic field directed into the screen (this is shown by X’s distributed evenly around the field – for the same reason as we draw current flowing into a screen with an X. If the field was coming out of the page there would be dots evenly distributed around the field). 

The charge will experience a force, which direction will the force be in the moment it enters the field?

vertically upwards

What happens next? If the force is acting vertically up it will accelerate upwards, could it then take the following path?

This will NOT be the path of the charged particle. Take the point labelled P, it is moving with a bearing of approximately 045^{\circ} . Using Fleming’s Left Hand rule again at this point, we can see that the force now changes direction. It is now pointing in the direction shown (with a bearing of 275^{\circ} ).

This means that the particle will take a circular motion. The force vector will always be perpendicular to both the velocity vector and that of the magnetic field, so as the particle changes direction, so will the force – at the same angular rate:

What do you think will happen if the particle above was negatively charged instead of positively and why?

Electron flow is the opposite to conventional current, this means that the force will act on the charge in the complete opposite direction. So initially the force will exert the charged particle downwards, it will the cause it to undergo circular motion in a clockwise direction (unlike the positively charged particle above which took an anticlockwise path.

Since it is normally electrons or protons that are fired into uniform magnetic fields, both or which have the elementary charge e = 1.6 \times 10^{-19} C the equation F = Bqv is often written as;

F =Bev

Now we are aware that these charged particle undergo circular motion, this force equation can be linked to that of circular motion:

F = Bqv      (1)

F = \frac{mv^{2}}{r}     (2)

Equating equations (1) and (2) gives:

Bqv = \frac{mv^{2}}{r}

This can be simplified to give:

Bq = \frac{mv}{r}

This is often rearranged to give a charge-to-mass ratio equation:

\frac{q}{m} = \frac{v}{Br}

This equation is particularly useful in pieces of equipment such a mass spectroscopy. For a given magnetic flux density (this can be controlled) and a given initial speed of a charged particle. The charge-to-mass ration is inversely proportional to the radius:

\frac{q}{m} \propto \frac{1}{r}

If not the charge-to-mass ratio, scientists use the mass-to-charge ratio: \frac{m}{q} \propto r . The following diagram show how different ratios of mass to charge will affect the radius teh particle will have. Complex computers can measure these radii and work backwards to determine the charge-to-mass ratio.

As can be seen at the bottom of this image, the atom that is inserted into the machine is first ionised. This means that it is stripped of an electron and therefore has an overall charge of +1.6 \times 10^{-19} C . If the charge is now known and the radius can be determined, the mass of a single atoms can be found. The mass of a hydrogen atom is 1.674 \times 10^{-27} kg , the mass of a helium atom is 6.646 \times 10^{-27} kg , this technique allows these masses to be distinguishable!

Further to this, it takes a high potential difference across an anode and cathode to accelerate charged particles to any user defined speed. The following is an equation which links this potential difference to the charged-to-mass ratio, ignoring the need for velocity altogether:

\frac{q}{m} = \frac{2V}{B^{2}r^{2}}

This is derived by starting with the force equation and equating it to circular motion (equation equations (1) and (2)):

Bqv = \frac{mv^{2}}{r}

Simlifying this and rearranging for v gives:

v = \frac{Bqr}{m}

Squaring both sides gives:

v^{2} = \frac{B^{2}q^{2}r^{2}}{m^{2}}      (3)

The charged particle would be accelerated across a potential difference, V .

This potential difference will apply work on the charged particle equal to W = Vq (as per the definition of potential difference).

Since the particle will gain kinetic energy, this equation becomes:

Vq = \frac{1}{2}mv^{2}

Rearranging this for v^{2} gives:

v^{2} = \frac{2Vq}{m}      (4)

Equating equations (3) into equation (4) gives:

\frac{2Vq}{m} = \frac{B^{2}q^{2}r^{2}}{m^{2}}

Simplifying this gives:

2V = \frac{B^{2}qr^{2}}{m}

Rearranging this so it is a charge-to-mass ratio and we get:

\frac{q}{m} = \frac{2V}{B^{2}r^{2}}

This is now an equation which would allow scientists to determine the charge-to-mass ratio based on the radius of curvature by controlling the potential difference between the anode and cathode during the acceleration of the charged particle.