Magnetic Fields Tutorial 1 - Magnetic Fields



Magnetic Fields

Domain Theory of Magnetism

Magnetic effect of an electric current

Magnetic field of a Solenoid

Field Strength of a Current

Force on a Current Carrying Wire

Direction of the Force

Required Practical



Magnetic Fields

You will be familiar with the basic notion of a magnetic field, in which magnetic materials experience a magnetic force. The force can come from an electric current, or a piece of magnetised material like a permanent magnet. Magnetic fields are different to other force fields because they are formed by dipoles (i.e. all magnets have a North and a South pole).




Electric fields are made using a single positive or negative point charge, and gravity is caused by point masses.  You never get a magnetic monopoles.  If you break up a magnet, you still get north and south poles:

This happens even if you grind the magnet down to atom sized particles.


It is worth revising some of the basic ideas that you will have come across in early secondary school.





Only iron, cobalt, and nickel and their alloys are magnetic.


Note that these elements are next to each other in the periodic table.  They are transition elements.


We can show the fields of two magnets attracting:


And repelling:



There is a neutral point where there is zero force.



Domain Theory of Magnetism

Magnets are thought to result from the action of tiny atomic magnets called domains.  This can be explained by the movement of electrons that represent a tiny electric current that results in magnetism.  In most materials, the currents cancel out.   When a magnetic material is unmagnetised, the domains are all jumbled up



If some of the domains are lined up, then the material is partially magnetised



If the domains are fully lined up, the magnet is saturated, and cannot be magnetised further.



Some materials like soft iron lose their magnetism quickly.  These are used for temporary magnets.  Permanent or hard magnetic materials do not lose their magnetism.




Magnetic effect of an electric current

Electric currents are always associated with magnetic fields.  The domains in a magnet are caused by the movement of electrons in shells.  Electric currents always produce a magnetic field, even if the wire itself is not made of a magnetic material.  The magnetic field of a single current carrying wire is like this:


The direction of the current is determined by the Screwdriver Rule.


The magnetic field strength depends on two factors:

The relationship is:


Note that the magnetic field strength varies inversely with the distance, while gravity and electric fields vary inversely with distance2.


Magnetic fields will interact.  In the picture below two current carrying wires with the currents flowing in the same direction.

There is a neutral point between the two wires where the magnetic field cancels:



The two wires attract.


If the currents are in the opposite direction, there is repulsion:


Notice that there is a resultant magnetic field.


This brings us onto a very important property of magnetic fields.  Magnetic field strength or flux density is a vector quantity.  The direction is important.  Note also that magnetic fields are three-dimensional, although we show them as two dimensional as it's easier to do this.  (I am not a very good artist.)


The SI unit for current is the Ampčre (amp) is defined in these terms:


1 amp is the constant current, which when maintained in two parallel conductors of infinite length and negligible cross-section that are 1 metre apart in free space, produces a force of magnitude 2 × 10-7 newton per metre along their length.


The equation for the force between two parallel conductors carrying currents I1 and I2 respectively that are r metres apart is this:



The term F/Dl is the force per unit length.  The term m0 ("mu-nought") is a constant called the permeability of free space.  The units for the permeability of free space are Henry per metre (H m-1)


m0 = 4p × 10-7 H m-1 = 1.257 × 10-6 H m-1

 So we can get a value for F/Dl:

F/Dl = (4p × 10-7 H m-1 × 1.0 A × 1.0 A) ÷ (2 × p × 1.0 m)


This gives us:

F/Dl  = 2.0 × 10-7 N m-1



Magnetic field of a Solenoid

A solenoid is a coil of wire usually wrapped around a former.



The magnetic field of a solenoid is like a bar magnet.




The diagrams show a three dimensional picture in two dimensions. We can show the directions of the current more easily using dot and cross diagrams:



The current is shown vertical to the page (or screen - let's get up to date!).  Outside the solenoid, the magnetic field is like this.



If the current goes clockwise, we get a south pole. If the current goes anti-clockwise, it's a north pole.



The Magnetic Field Strength of an Electric Current (Welsh Board and Eduqas)

Consider a solenoid of n turns per metre which is carrying a current of amps.


Note that the term n is turns per metre.  So you need to divide the total number of turns by the length.


If we measure the flux density in the solenoid well away from the ends, we find that:

µ I


µ n


So we can write:

µ nI


There is a constant of proportionality which is called the permeability of free space.  It given the physics code m0 ("mu-nought" - the symbol 'm' is 'mu', a Greek lower case letter 'm').  The units for the permeability of free space are Henry per metre (H m-1)


m0 = 4p × 10-7 H m-1 = 1.257 × 10-6 H m-1


Do not mix up the permeability of free space m0 with the permittivity of free space e0.  The two words sound similar.


Strictly speaking, the permeability of free space applies to a vacuum.  However the value in air is very similar.


In some calculators, keying in "4p ×10(-)7" gives a syntax error.  Key in "4p × 1 ×10(-)7"  or "4 ×10(-)7 × p and it will work.


We can write an equation for this:

B = m0nI


Worked Example

A solenoid of length 2.5 cm has 200 turns. A current of 1.65 A flows through it.

Calculate the resulting magnetic field strength. Give your answer to an appropriate number of significant figures.


The number of turns per metre = 200 turns ÷ 0.025 = 8000 turns per metre.

B = 4p × 10-7 H m-1 × 8000 m-1 × 1.65 A = 0.0166 T = 0.017 T (= 17 mT)

2 significant figures as the length is to 2 s.f.


If we slide a bar of magnetic material into the central space of the solenoid, we have added a core.  The magnetic field strength is increased by a factor mr ("mu-arr"), which is called the relative permeability.  This factor is a ratio; therefore there are no units.  Our equation becomes:


B = m0 mrnI


The relative permeability of pure iron is about 5000.  Non-magnetic materials have a relative permeability of 1.



Force on a Current Carrying Wire

You will be familiar with the motor effect.  If we put a current carrying wire in a magnetic field, we see that there is a force.  The picture below shows a typical demonstration:



A carbon rod is placed in the magnetic field of a large permanent magnet.  (You can't get anything less magnetic than carbon!) The brass rails connect the carbon rod to the power supply.  When a current flows through the carbon rod in the directions indicated by the arrows, the rod experiences a force and moves from left to right.  This shows that the current in the carbon rod makes a magnetic field that interacts with the magnetic field from the permanent magnet to produce a force.


If we turn the magnetic field so that it is parallel to the rails, the force is zero.




Direction of the Force



As we pass a current through the wire, there is a force that acts on the wire at 90o to the direction of the magnetic field.  This is given by Fleming’s Left Hand Rule with which you will be familiar.



We can work out the force that is exerted on the wire quite simply.  Experiment shows us that the force is proportional to:

This is summed up in a simple formula:


 F = BIl


[B magnetic field strength; I – current in A; l – length in m]


The term B is called the magnetic field strength, or the flux density, and is measured in Tesla, T.  Flux density is a vector quantity  The magnetic flux density can be thought of as the concentration of field lines.  We can increase the force by increasing any of the terms within the equation.  If we coil up the wire, we increase its length within the magnetic field.



Worked Example

A current of 8.5 A flowing through a magnetic field is found to exert a force of 0.275 N.  The length of wire in the magnetic field is 5 cm.  What is the value of the magnetic field?


Formula first:  F = BIl


Ţ B =  F  = ___0.275 N ___ = 0.647 T

         Il       8.5 A ´ 0.05 m


Question 1

In a demonstration of the above equation, the length of wire in a magnetic field is 0.05 m.  When a current of 2.5 A flows, a force of 0.01 N is shown.  What is the magnetic field strength?



If the field is at any angle other than 90 degrees, the formula takes this into account with the sine function:


Question 2

In a demonstration of the above equation, the length of wire in a magnetic field is 0.05 m. When a current of 2.5 A flows, a force of 0.01 N is shown. What is the magnetic field strength if the wire is at an angle of 35o to the field?




Required Practical - Force on a wire from a magnetic field

A simple experiment can be carried out to measure the force produced when a current flows through a magnetic field.  The apparatus set up is straight-forward:



In the diagram, a retort rod is used.  A wire in a glass tube is also effective.  The horseshoe magnets are from the motor kits that most schools and colleges have.


You need to:


You then need to plot a graph which will show direct proportionality:



The magnetic field strength will come from the gradient.


Gradient = Bl


Therefore we divide the gradient by the length to get a value for the magnetic flux density.