Tutorial 3 - Magnetic Force

Learning Objectives

To learn about the field of a single wire;

To study the motor effect;

To analyse the forces between current carrying wires;

To calculate the force from an electromagnet.

Key Questions

How do we work out the magnetic field of an electric current?

How do we work out the force on a wire in a magnetic field?

How do we work out the force between two wires?

What is the force from an electromagnet?

Magnetic Force Fields

When you did electric fields and gravity fields at Physics A-level, you will have studied radial and uniform fields.  In magnetic fields, it is not possible to make a truly radial field, although it is possible to make a magnet so that the field lines are nearly radial.  In an electric field, you would get a radial field from a point charge that is either positive or negative.  You did calculations on test charges. In a gravity field, you get a radial field that is always attractive as the result of a point mass.  You did calculations on test masses.  


In a magnetic field, we cannot have a radial field from a monopole, as magnets are always dipoles.  We cannot have test poles, as these would need to be monopoles, which we cannot have.  Magnetic fields are secondary fields, as a result of the flow of a current.


Magnetic fields do not follow the inverse square law.


In the examples discussed in this tutorial, we will assume that all fields are uniform, i.e. the field lines are all equally spaced.


There are magnetic field lines that bulge out on either side of the magnet.  The effect of these can generally be ignored.

The Magnetic Field of an Electric Current

In Tutorial 1 we saw that the magnetic field around a wire is circular, with its direction determined by the screwdriver rule.  In this case the current is coming out of the screen towards you.  The field lines are oriented anticlockwise.

The flux density depends on two factors:

  • the current;

  • the distance from the wire.

The relationship is:

From this you can see that:

This is not the same as force fields like gravity or electric fields, where the strength varies as 1/r2.


Question 1

A flux density of 3.5 mT is measured 15 cm from a current carrying wire.  Calculate the current in the wire.

m0 = 4p × 10-7 H m-1



The Motor Effect

You may remember a demonstration like this when you did your A-level.  (If you are doing Access L3, your tutor will demonstrate this.)

The carbon rod is not magnetic.  In fact you would not want a magnetic material, as it would be attracted to the magnet (and would be quite difficult to remove).   However the field made by the current interacts with the field of the magnet.


A current carrying wire in a magnetic field experiences a force that is perpendicular to the current and perpendicular to the field.  The direction of movement is determined by Fleming's Left Hand Rule.


The greatest force is when the current is at 90o to the magnetic field.  There is zero force when the wire is parallel to the field.

In this case, there is zero force acting on the rod, as the field is acting at 0o to the rod.


The force F (N) depends on four factors:

  • The flux density, B (T);

  • The current, I (A);

  • The length, l, (m);

  • The angle the wire makes to the field (o).

The equation is:

If the angle is 90o then sin q = 1, and the relationship is simpler:

F = BIl


A simple experiment demonstrates the idea:


Question 2

In the experiment above, a current of 5.5 A causes the reading on the balance change by 6.3 g.  Each of the three horseshoe magnets is 5.0 cm long.  Calculate the flux density of the magnets.


The graph looks like:

The gradient is used to work out the flux density:

Flux density (T) = gradient (N A-1) ÷ length (m)


The force between two current carrying conductors

In Tutorial 1 we looked at the magnetic fields of two parallel wires carrying currents in the same direction and in opposite directions.

Let's have a look at the effect of the magnetic field of the left hand wire on the right hand wire.  For the currents going in the same direction we see:

In this case the current is coming out of the screen towards us.  Therefore, if we apply Fleming's Left Hand Rule to Wire B, we see that it moves to the left:

If we do the same thing to Wire A using the field from Wire B, the field is coming downwards.  Therefore the motion is to the right.  Therefore the two wires will attract.


The force of attraction depends on:

  • The current in Wire A, IA, (A);

  • The current in Wire B, IB, (A);

  • The length of the wires, l, (m);

  • The distance between the wires, r, (m).

The currents may, or may not, be the same.  The length of the wires is the length of where they are parallel to each other.  We know that the value of the force on Wire B is given by:

Wire A is providing the magnetic field.  We also know that the flux density from wire A is given by:

Therefore we combine the two to give us:

This is quite a helpful result, because we get the p terms cancelling out, which makes the numbers easier.


Question 3

Two parallel wires, X and Y, of length 0.8 m have currents 3.0 A and 5.0 A flowing through them respectively.  The two wires are 5.0 cm apart.  Calculate the force between them.

m0 = 4p × 10-7 H m-1


Question 4

A coil is connected to a 50 Hz AC supply.  Explain why you can hear it buzzing.  What frequency is the buzzing?  Explain your answer.



If the currents are going in the opposite directions, the wires will repel.  The current in Wire B is going into the screen.

If we apply Fleming's Left Hand Rule, we see:

Wire B moves to the right.  Wire A will move in the direction it did before, i.e. to the left, as the field is still vertically upwards.  Therefore the two wires repel.  The value of the repulsive force is still given by the relationship:

Ampère's Law

The ampere is defined in terms of the force between two current carrying wires.

Consider two wires of infinite length 1 m apart carrying a current of 1.0 A:


The force on each metre length of the wire is 2 × 10-7 N m-1 when the current is 1.0 A.


Question 5

Show that the statement above is true.


Ampère's Law states that:

One ampere is the constant current which, when passing through two infinitely long parallel straight conductors of negligible cross-section placed 1 metre apart in a vacuum, produces a force between them of 2 × 10-7  N per metre of their length  between them.

Force due to an electromagnet

A simple experiment can be carried out to measure the force from an electromagnet.  Slotted masses are attached to a fitting made of a magnetic material that will cover a core of magnetic material that is placed inside a current carrying coil. 

A value of current is selected, and  masses are added until the slotted masses fall off.  The maximum mass held is recorded, and converted to a force by multiplying by 9.81 N kg-1.  Then another current is selected.


We know that for a coil with a core in it that the flux density, B, is given by:

  • n is the number of turns per metre.

  • I is the current (A).

The force F is related to flux density by this equation:

  • A is the area (m2) of the core which we can easily work out by measuring the core.

  • This is true for any magnet, whether it's a permanent magnet or an electromagnet.

Do not use the relationship F = BIl for this situation.


The relationship above is for a wire carrying a current at 90o to the magnetic field.




We can substitute the first equation into the second which gives us:

It doesn't take a genius to see that some cancelling can be done to give:

This relationship has the following assumptions:

  • The field within the core is uniform;

  • The fringe effects (magnetic field lines leaking away from the core) are negligible.


Worked Example

A coil has 1200 turns and a length of 7.5 cm.  It is of rectangular cross-section, 2.0 cm × 3.0 cm.  A direct current of 0.75 A is flowing through the coil.  The core fits the coil tightly and has a relative permeability of 500.  Calculate the force it produces.

m0 = 4p × 10-7 H m-1


Work out the turns per metre:  n = 1200 ÷ 0.075 m = 16000 m-1

Work out the area: A = 2.0 × 10-2 m × 3.0 × 10-2 = 6.0 × 10-4 m2


Now we can do the calculation:

F = 27 N (2 s.f.)


The result shows why even a few turns of wire around a nail makes quite an effective electromagnet.  You may well have done a simple experiment in which you wrapped a few turns of copper wire onto a nail and saw how many paper clips you could pick up.


Question 6

A solenoid consists of 100 turns of wire wrapped around a former that is 5.0 cm long.  It has a circular cross-section of diameter 1.0 cm.   A core of magnetic material is inserted into the solenoid so that it fits tightly.  It forms an electromagnet that can hold a weight of 3.7 N when a current of 0.85 A is flowing through it.   Calculate the relative permeability of the material.



This relationship is really an approximation in which the assumptions were that:

  • The field was uniform throughout the core;

  • The leakage of field lines was negligible.

If the relative permeability is high, then the relationship is correct.  If the relative permeability is low, then there is a correction factor that can have a significant effect on the result:

For a material of relative permeability 5000, the correction factor would have an insignificant effect.  If the material were only slightly magnetic, then the correction factor would be much more significant.


Now have a go at the multiple choice test.