Magnetic Fields

Key Words

Magnetic, Flux, Flux Density, Flux Linkage

In the exam you are expected to know about:

  • understand that the force is dependent on the current, the length, and the magnetic field strength;

  • define the Tesla as unit for magnetic field strength;

  • use the formula F = BIl;

  • recall Fleming's Left Hand rule;

  • understand the relationship F = BQv. 

  • explain how this gives rise to a circular path. 

  • understand an application in Physics for this, for example a cyclotron.

  • be able to use the equation F = BA when B is normal (perpendicular) to A

  • understand Faraday’s and Lenz’s Laws for a flux change at uniform rate. 

  • be able to use the relationship E = NDF/Dt.

  • Emf induced in a coil rotating uniformly in a magnetic field: E = BANw sin wt

  • The transformer effect

  • The transformer equations: Nprim/Nsec = Vprim/Vsec and Nprim/Nsec = Isec/Iprim;

  • The transmission of electricity at a high voltage.

 

 

 

Force On A Current Carrying Wire In A Magnetic Field

You will be familiar with the basic notion of a magnetic field, in which magnetic materials experience a magnetic force.  However it is worth revising some of the basic ideas that you will have come across in early secondary school.

 

Revision

 

     

 

 

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.

 

 

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 ` is called the magnetic field strength, or the flux density, and is measured in Tesla, T.  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.

Þ 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?

ANSWER


This relationship holds true as long as the current is at 90o to the magnetic field.  If the wire is at an angle to the field, the relationship takes this into account by changing to:

 

F = BIl sin q

 

 

Force on a Charge

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]

 

The charge is usually the electronic charge, 1.6 ´ 10-19 C.

 

Question 2 

An electron accelerated to 6.0 ´ 106 m/s 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?

ANSWER

 

 

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:

 

Therefore:        

                                                           

 

This rearranges to give us:

 

 Worked Example

An electron passes through a cathode ray tube with a velocity of 3.7 × 107 m/s.  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?

F = Bqv and F = mv2/r Þ r = mv

   Bq

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

0.47 × 10-3 T × 1.6 × 10-19 C


 
Question 3

In a particle physics experiment, a detector is place in a magnetic field of 0.92 T.  A particle is found to produce a track of radius 0.5 m.  Other experiments have shown that the particle carries a charge of +1.6 ´ 10-19 C and that its speed was 3.0 ´ 107 m/s.  What is the mass of the particle?  How does it compare to the mass of an electron (9.11 ´ 10-31 kg)? 

ANSWER

 

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. 

 

 

 

Þ rearranging gives us

 

           

 

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

 

 

The voltage will change twice in each alternating current cycle, therefore in the period of one cycle, the time is 2t.  Since f = 1/T, we can write the expression as:

 

 


 

Magnetic Flux Density, Flux, and Flux Linkage

We have seen how B is called the magnetic field strength, or the flux density, and is measured in Tesla, T.  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.

 

If we look at the magnetic field of a solenoid, we know that it is like a bar magnet:

 

 

We can see that the magnetic field strength is uniform within the solenoid.  However the flux density becomes less at the ends, as the field lines get spread out.

We need a term that tells us the number of field lines, and it is called the magnetic flux.  It is given the physics code F (‘Phi’, a Greek capital letter ‘Ph’), and has the units Weber (Wb).  The formal definition is:

The product between the magnetic flux density and the area when the field is at right angles to the area.

In code we write:

F = BA

Remember that flux density is the number of field line per unit area, not unit volume!

The flux linkage is the flux multiplied by the number of turns of wire.  If each turn cuts (or links) flux F, the total flux linkage for N turns must be NF.  We can also write this as NBA.  In other words:

Flux linkage = number of turns of wire ´ magnetic field strength ´ area

Question 4

How much flux links a 200 turn coil of area 0.1 m2 when it is placed at 90o to a magnetic field of strength 2.5 ´ 10-3 T?

ANSWER

The flux linkage can be changed in two ways:

We give the change in flux linkage the physics code DF.

Question 5

The coil in question 4 is now turned so that it makes an angle of 60 o with the magnetic field lines.  What is the change in flux linkage?

ANSWER

 

Electromagnetic Induction

If we pass a current in a wire in a magnetic field, we know that the wire will move.  It is therefore reasonable to suppose that if we move the wire in a magnetic field, and the wire is connected to an outside circuit, a voltage and current are induced.  If the wire is not connected, a voltage only is induced.  Consider this demonstration:

If we move the magnet parallel to the wire, the galvanometer hardly responds.  However, if we move the magnet across the wire, then we see a definite reading on the galvanometer.  The current (and voltage) induced on a single wire is rather small, but is increased by having more turns of wire.  For any voltage to be induced, we must move the magnet.  We call this voltage the induced electromotive force.  It is often given the code Î, a fancy letter ‘E’.

 

Faraday’s Law and Lenz’s Law are two important rules that govern this effect.

 

Faraday’s Law is a formal definition of the effect:

            The induced e.m.f. across a conductor is equal to the rate at which flux is cut.

 

Lenz’s Law says:

  The direction of any induced current is such as to oppose the flux change that caused it.

 

The induced e.m.f. sets up a current that would oppose the force that is pulling the wire.  If the force were to assist the motion, we would get acceleration, and an increase in kinetic energy.  This would break the Law of Conservation of Energy.  In other words, we cannot get something for nothing.

 

Lenz’s Law is important in motors and generators.  As a motor speeds up, it acts like a generator to produce a back e.m.f. to oppose the current flowing in the motor.  Therefore the current through a fast running motor is quite small.  When it is running slowly, a big current flows.  Electric motors are therefore very suited to railway use, where big currents are needed to get trains moving, and there is no need for a gearbox that is needed for a diesel engine.

 

The effect that we have seen is summed up in the relationship:

                       

 

[N - number of turns; E– e.m.f (V); DF/Dt – rate of change in flux (Wb/s)]                         

                                  

Worked Example

A single turn of wire of cross-sectional area 5.0 cm2 is at 90o to a magnetic field of 0.02 T, which is reduced to 0 in 10 s at a steady rate.  What is the e.m.f. induced?

Two formulae to use:  F = BA            and             Î = - NDF   

     Dt

We need to work out the flux:

                        F = BA = 0.02 T × 5 × 10-4 m2 = 1 ×10-5 Wb

Now we can work out the e.m.f:

                       Î = - NDF = 1× 1 ×10-5 Wb = 1 ×10-5  V

      Dt           10 s

 

Question 6

A search coil has 2500 turns and an area of 1.5 ´ 10-4 m2. It is placed between the poles of a large horseshoe magnet.  It is rapidly pulled out of the field in a time of 0.30 s.  A datalogger records an average value for the emf of 0.75 V.  What is the flux density between the poles of the magnet? 

ANSWER

 

 

We can use a magnetic field to induce a voltage in two ways:

 

1.      Relative movement.  The size of the voltage depends on:

2.      Changing a magnetic field.  We don’t have to make the magnetic field move.  If we turn the current on or off, there is a change in the magnetic field, and that induces a voltage in a second unconnected coil.  This is called the transformer effect or mutual induction.

 

The transformer is a machine that is simplicity itself.  It consists of:

 

The two coils are electrically completely different circuits.  Either of the coils can act as a primary.  The laminated core is made up of layers of soft iron separated by an insulating layer of varnish or glue.  This reduces losses from eddy currents.  Soft iron is NOT soft like putty; it is heavy and hard.  However "soft" means that it loses its magnetism immediately the current is turned off.  Therefore the magnetic field can change forwards to backwards as the current changes.

 

The ratio of the input voltage to the output voltage is the same as the ratio of the number of turns on the primary to the number of turns on the secondary.  We can write this as:

 

Number of turns on the primary    = Primary Voltage

Number of turns on the secondary    Secondary voltage

 

In Physics code:

 

N1 = V1

N2    V2

 

If N1 is greater than N2, we have a step-down transformer, because the voltage is reduced.  A step-up transformer increases the voltage.

 

If a transformer is 100 % efficient (and it nearly is) we can say that:

 

 power in = power out

 

V1I1 = V2I2

 

Therefore we can say that when the voltage is lower, the current is bigger.  We can rewrite the transformer equation in terms of current to give us:

 

N1 = I2

N2    I1

 

In practice, the transformer is about 97 % efficient.  When a large transformer is transferring a lot of energy, even 3 % losses produce a fair amount of heat.  Therefore the transformer is cooled with oil which is pumped to heat exchangers.

 

 

The picture above shows the huge transformers used at a power station.  You can see the massive cables coming out from the generator (they are like pipes).  In this power station, the generator voltage is 15 000 V, and this transformer steps the voltage up to 275 000 V.

 

Question 7

A power station generator generates 500 MW at a voltage of 15 000 V.

(a) What is the current?

(b) The voltage is stepped up to 275 000 volts.  Assuming that the transformer is 100 % efficient, what is the current in the secondary?

(c) What is the turns ratio in the transformer?

Answer

 

 

Transmission of Electricity

To carry the kinds of currents you worked out in Question 7, you would need very thick wires.  Heavy currents make even thick wires get hot.  The power lost is worked out using:

P = I2R

 

Question 8

A low voltage transmission line is carrying a current of 30 000 A.  Over its whole length, the resistance of the transmission line is 1.5 W.  How much power is lost?

Answer

 

Your answer would be about the whole output of a big power station.  That energy would be lost as heat, simply to warm up the countryside.  So electrical energy is distributed at very high voltages, and relatively low currents.

 

Question 9

A high voltage transmission line is carrying a current of 1000 A.  Over its whole length, the resistance of the transmission line is 1.5 W.  How much power is lost?

Answer

 

You can see that the lost power is much less.

 

In power stations, alternating current is generated typically at 25 000 V (with a current of 30 000 A) from each alternator (ac generator).  The alternators are connected by short and massive cables to a step up transformer immediately outside the building.  The voltage is stepped up to 275 kV.  Much thinner cables can carry the electricity to where it's needed, using a network of high voltage transmission lines called the National Grid.  The transmission lines are carried by transmission towers (pylons) to substations where the voltage is reduced by step-down transformers:

While such high voltages are potentially extremely dangerous, the distribution from large power stations is much more efficient that lots of small local power stations, so less fuel needs to be burned overall.

 

Summary

Force on a current carrying conductor:

                                                   F = BIl

 

 

Force on a charged particle:

                                                      F = Bqv

 

Trajectory of a Charged Particle in a magnetic field:

                                                      v = Bqr

                                                            m

 

 

Flux Density

Magnetic field strength, B.  “Concentration of field lines”

 

Flux 

Product of field strength and area.  F = BA.  Total number of field lines.

 

Flux Linkage

Product of flux and the number of turns

 

Faraday’s and Lenz’s Laws

                                                             E = - NDF     

                                                                         Dt

         

The minus sign tells us that the emf opposes the change.

 

Transformer Equation

                                                            Nprim = Vprim = I sec

                                                            Nsec      Vsec      Iprim

                                   

Now try the Topic Quiz

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