Tutorial 2 - Magnetic Quantities

Learning Objectives

To review and define the terms used in electromagnetism.

To make quantitative links between electrical currents and magnetism.

Key Questions

What are the terms used in magnetism?

How is an electric current linked with a magnetic field?

Magnetic Quantities

If we look at a magnetic field, we find that the field lines are close together near the poles of the bar magnet.  Here the field strength is high.  If we move away from the ends of the magnet, the field lines are further apart, so the magnetic field strength is less.  Look at the picture:

The weak magnetic field has three field lines going through a wide circle.  The magnetic field strength has a low value.  The strong magnetic field has seven field lines going through a circle of much smaller diameter.  The lines are more concentrated.  The magnetic field strength has a much higher value.

 

The total number of field lines is referred to as magnetic flux.   It is defined as the number of field lines passing through a surface.  It has the physics code F ("Phi", a Greek upper case letter 'F' or 'Ph').  The units are Weber (Wb).

 

The magnetic field strength is referred to as flux density.  It is a vector quantity, so has a direction as well as a value.   It is defined as the flux per unit area.    This definition gives the equation:

The physics code is B and the units are Tesla (T), or Weber per square metre (Wb m-2).  1 T = 1 Wb m-2

 

The flux, F, is given by rearranging the equation:

Question 1

A magnetic field has a strength of 55 mT.    The field is concentrated into a rectangular area of dimension 5.0 cm × 3.0 cm.  Calculate the flux.

Answer

The Magnetic Field Strength of an Electric Current

Consider a solenoid of n turns per metre which is carrying a current of I 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:

B µ I

and

B µ n

Question 2

Why is the magnetic field strength measured on the inside of the solenoid, away from the ends?

Answer

 

The direct proportionality is shown by these graphs:

So we can write:

B 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:

Question 3

Describe how you would would work out the constant of proportionality using the current graph above.

Answer

Question 4

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.

Answer

 

In some textbooks this idea is approached using the magnetomotive force.  It is given the physics code Fm and the units are ampere turns.  Since turns is dimensionless, it is the same as amp (A).  The equation is given as:

Fm = NI

In this context the magnetic field intensity (H) is given in this equation:

Magnetic field intensity has the units amp per metre (A m-1).

 

The term N/l is the term turns per metre.  Notice that a different physics code for magnetic field intensity is used here, because it has not taken into account the permeability of free space.  The argument then develops the idea of the ratio B/H as a constant, i.e.

 

In the end it all ends up the same.

Question 5

Show that the equations used in this alternative approach give a consistent end result, B = m0nI

Answer

Linking with Flux

Since

we can write an equation that tells us the flux from a solenoid:

Question 6

A solenoid has 150 turns wound over a former that is 10 cm long and is of uniform circular cross-section diameter 3.0 cm.  A current of 5.6 A flows through it.  Calculate the flux that it produces.

Answer

Question 7

The same solenoid as above is needed to produce a magnetic field of magnetic field strength 0.4 T.  Calculate the current needed to produce the field of this value.

Answer

Adding a Core to a Solenoid

The answer to Question 6 showed that a heavy current would be needed to make magnetic field of the coil as strong as that of a magnet commonly found in a school or college Physics laboratory.  The current would probably burn the coil out.  Thus electromagnetism would be just a physics curiosity instead of something really useful.  We can strengthen the magnetic field considerably by adding a core.  The core needs to be made of a magnetic material.  If we put in a core of non-magnetic material, there is no change in the magnetic field strength.

 

The increase in the magnetic field strength depends on the material.  The factor by which the magnetic field strength is increased is called the relative permeability.  It is given the physics code mr ("mu-arr") and there are no units; it's just a number.  The definition or relative permeability is:

For a vacuum, mr = 1.  The relative permeability for air, mr = 1.00000037, so it's the same.  All non-magnetic materials have a relative permeability of about 1.0; it's only when we go to several significant figures that tiny variations are seen.  In these notes, all non-magnetic materials will be considered to have mr = 1.00.

 

The equation that links magnetic field strength to current in a solenoid now becomes:

The product of the permeability of free space and relative permeability is the absolute permeability or (simply) permeability.  It is give the physics code m and the units are Henry per metre (H m-1).  The equation is:

 

The table below shows some values for different non-magnetic and magnetic materials:

 

Material Permeability µ [H m-1] Relative permeability μ/μ0
Air 1.25663753×10−6 1.00000037 
Aluminum 1.256665×10−6 1.000022
Austenitic stainless steel 1.260×10−6 - 8.8×10−6 1.003–7 
Bismuth 1.25643×10−6 0.999834
Carbon Steel 1.26×10−4 100
Cobalt-Iron (high permeability strip material) 2.3×10−2 18000
Concrete (dry)   1
Copper 1.256629×10−6 0.999994
Electrical steel 5.0×10−3 4000
Ferrite (manganese zinc) >8.0×10−4 640 (or more)
Ferrite (nickel zinc) 2.0×10−5 – 8.0×10−4 16–640
Ferritic stainless steel (annealed) 1.26×10−3 - 2.26×10−3 1000–1800
Hydrogen 1.2566371×10−6 1
Iron (99.8% pure) 6.3×10−3 5000
Iron (99.95% pure Fe annealed in H) 2.5×10−1 200000
Martensitic stainless steel (annealed) 9.42×10−4 - 1.19×10−3 750–950
Martensitic stainless steel (hardened) 5.0×10−5 - 1.2×10−4 40–95
Metglas 2714A (annealed) 1.26×100 1000000
Mu-metal 2.5×10−2 20000
Mu-metal 6.3×10−2 50000
NANOPERM® 1.0×10−1 80000
Neodymium magnet 1.32×10−6 1.05
Nickel 1.26×10−4 - 7.54×10−4 100 – 600
Permalloy 1.0×10−2 8000
Platinum 1.256970×10−6 1.000265
Sapphire 1.2566368×10−6 0.99999976
Superconductors 0 0
Teflon 1.2567×10−6[10] 1
Vacuum 4π × 10−7 (µ0) 1, exactly
Water 1.256627×10−6 0.999992
Wood 1.25663760×10−6 1.00000043

Data from Wikipedia

 

Question 8

A solenoid of 1500 turns per metre is needed to produce a magnetic field of magnetic field strength 0.4 T.  A piece of 99.8 % pure iron is placed within the solenoid.  Calculate the current needed to produce the field of this value.

Answer

 

You can see from your answers to questions 7 and 8 that the current needed to produce a magnetic field of 0.4 T is much less with the iron core than the value without the iron core.  The graph below shows a flux density against current graph for no material, a iron with a permeability of 5000, and cobalt-iron with a permeability 18000.

Notice that the graphs for the cores are not linear, but the gradient decreases.  This is because the materials become saturated, as all the domains line up.  The graphs are linear when the currents are low.  However the model works for when the currents are low, which is most of the time.

Magnetic Field of a Flat Coil

Consider a flat coil of N turns, and radius r metres.  It is carrying a current of I amps.  The current is going clockwise as we look at it.

The magnetic field looks like this:

As the coil is circular, all the vector components of the flux density add up at the centre, where the maximum magnetic field is measured.

 

The flux density at the centre is directly proportional to the current, I and the number of turns, N.  It is inversely proportional to the radius, r, of the coil. 

 

The formula for the flux density at the centre is this:

 

Question 9

Calculate the flux density in the centre of a coil of 30 turns of diameter 10 cm when a current of 3.6 A is flowing in the coil.

Answer

 

The above calculation involved a point right in the middle of the coil.  What happens if we move a distance, z, along the axis of the coil? 

 

 

The flux density at z metres from the centre, Bz is given by a more complex looking equation:

 

Any term raised to the power 3/2 means it's the square root of the cube of the term.  For example 43/2 = Ö64 = 8.

 

Question 10

Calculate the flux density of a coil of 30 turns of diameter 10 cm at a point 7 cm from the plain of the coil when a current of 3.6 A is flowing in the coil.

Answer

The derivations of the equations above are quite involved, and will no doubt form part of a first year university lecture.

We will look at Helmholtz coils (two parallel flat coils a short distance apart) in a later tutorial.

Now try the multiple choice quiz.

Test

Home

Magnetism