4/4/2006 magnetic fields (© f.robilliard) 1flai/theory/lectures/magneticfields.pdf · 4/4/2006...

4/4/2006 Magnetic Fields (© F.Robilliard) 1

F

x

y

z

NS

+

v

+q θθθθ


Introduction:It has been known, since antiquity, that when certain pieces of rock are hung by a thread, from their centre of mass, they will align themselves in a North-South direction. These rocks were naturally-occurring magnets, composed of magnetite (Fe3O4). One end was called a North pole (N), and the other end, a South pole (S), depending on the geographic direction in which they pointed.

Like poles (N & N, or S & S) were found to mutually repel each other. Unlikepoles (N & S) were found to mutually attract.

This force of attraction, or repulsion, is the magnetic force. A region in which a magnetic force exists is called a magnetic field.

(The NS alignment of magnets, was due to magnet forces produced by interaction of the magnet, with the earth’s global NS magnetic field.)

Our first task must be to define, precisely, the magnitude and direction, of the magnetic field.

We do this in terms of the magnetic force.


Field Definition Using a Test Magnet:

N S

The direction of a magnetic field can be defined as the direction in which the SN poles of a small test magnet would align, when placed at a point in the field.

S->NDirection of the field is in the S-to-N direction of the test magnet.

However, defining the magnitude (strength) of the field in this way is problematic. The net magnetic force on a small test magnet in a uniform magnetic field is zero – the forces on N and S poles are equal and opposite.

Use of an isolated N pole would alleviate this difficulty, but single magnetic poles (monopoles) do not exist.


Field Definition Using a Charge:

When a charge is stationary in a magnetic field, no magnetic force acts on it.

A better way to define a magnetic field is in terms of the magnetic force that acts on a moving electrical charge

But when the charge moves, relative to the field, a magnetic force acts on the charge. This force depends on -

Because the magnetic force depends on the magnitude and direction of the magnetic field, it can be used to define the field.

- the charge, - the strength of the magnetic field- the magnitude and direction of the velocity of the charge,

relative to the direction of the field.


Experimental Observations:

F

x

y

z

Say a uniform magnetic field is directed in the +x-direction

+q

v+

We move a particle with positive charge, +q, through the field, with a velocity, v, in the +xz-plane, at an angle, θ, to the direction of the field.

Experiment shows, that a magnetic force, F, acts on the charge in the +y-direction – perpendicular to both the direction of the field, and the direction of the velocity, v.

The magnitude, F, of this force- is proportional to the charge, q, - is proportional to the (v sin θθθθ) component of v- is not affected by the (v cos θθθθ) component of v- depends on the strength of the field

Because the magnetic force, F, depends on the magnitude and direction of the magnetic field, it can be used to define the field.

θθθθ Let’s resolve the velocity, v, into- a (v cos θ)θ)θ)θ) component in the direction of the field, and - a (v sin θθθθ) component perpendicular to the direction of the field.

vcosθθθθvsinθθθθ


Definition of Field Vector B:

θθθθsin B vqF =

F

x

y

z

+

v+q θθθθ

Therefore the magnetic force, F, is perpendicular to both v, and B directions, as before.

Taking account of the experimental facts noted above, namely that the magnitude, F, of the magnetic force is proportional to the charge q, proportional to the (v sin θθθθ) component of the velocity of the charge (the component perpendicular to B), and following our assertion, that F should be proportional to the magnitude of B, we can write -

We represent the magnitude and direction of the magnetic field, at the location of the charge, by a field vector B.

B

We define the magnitude (or strength) of the magnetic field, B, so that the magnitude of the magnetic force, F, and the magnitude, B, of the field are in direct proportion.

This relationship defines the magnitude of the field vector B.

The direction of B is defined according to the relative directions in the above figure.


Vector Definition of B:

( ) BvB vF ×=×= q q

The definition of field vector B can be expressed more elegantly using the following cross-product expression, which incorporates magnitudes and directions of all vectors --

This expression should be thought of as a general implicit definition of the magnetic field vector, B, in terms of the magnetic force, F that acts on a charge, q, that moves with a velocity, v, at a particular point in a magnetic field.

Having thus defined the field vector, B, we can use the above expression to find both the magnitude, and direction, of the magnetic force, F, that would act on a moving charged particle, in a magnetic field.

SI Unit for B: B = F/(qv sin θ), therefore units for B will be -

N C-1 m-1 s = Tesla = T

A charge of 1 C would experience a magnetic force of 1 N if it moved perpendicular to a magnetic field of 1 T, at a velocity of 1 m/s.

Examples: Earth’s magnetic field: ~50 µT; conventional laboratory magnet: ~2 T


Along & Across the Field:

x

y

z

N S+ v+q

F=0When the charge moves in the direction of the field (θ=0), the magnetic force is zero.

As the angle, θ, between v, and B, changes, the magnetic force will vary as (sin θ).

B vq sin B vqF == θθθθ

F

x

y

z

+v

+q

When a charge moves perpendicularly to the direction of the magnetic field (θ = 90 deg), the magnetic force is maximum.

0 sin B vqF == θθθθ

The direction of the magnetic force is given by the Right Hand Rule (RHR) for cross products.


Example:An electron (q = -1.6 x 10-19 C) passing, with a velocity of 100 m/s, in a direction perpendicular to the magnetic field between the poles of a “horseshoe magnet”, experiences a magnetic force of 32 aN. What is the strength of the field?

N S

B-

v

F

( ) T 2.0 100 10 x 1.6

10 x 32

qvF

B thus

B vqF

19-

18-

===

=


Magnetic Flux Lines:

A magnetic flux line is a line drawn in a magnetic field with shape such that a tangent to the line at any point corresponds to the direction of the magnetic field vector B at that point.

As we have done with electric fields, magnetic fields can be represented graphically by flux lines.

Although the B-vector is the quantity that fundamentally represents the magnetic field, there is a more potent way to represent the field, graphically.

The strength of the field is represented by the closeness of flux lines drawn in a given region of the field – a strong field has close-together flux lines.

Any point, P, in a magnetic field, is characterised by a field vector B. Spaces which have a characteristic vector defined at every point, are called vector fields.


Examples of Flux Lines:

flux lines

N S

BN S

B

flux lines

Later we will develop a quantity called flux, to represent the density of flux lines.

Non-Uniform magnetic field –near a bar magnet:

Uniform field between the poles of a horseshoe magnet:


BvF ×= q ( )� ×=Q

P

i BdsF


Motion of a Free Charged particle in a Magnetic Field:We turn our attention firstly to the motion of charged particles that move freely, in a magnetic field.

Say a free charged particle, of mass, m, and charge, q, is projected into a uniform magnetic field, of vector B. How will it move?

Assume that v is perpendicular to the field vector, B,.

The magnetic force, F, on the particle, is given by

F = q (v x B)

(Magnitude of F) = F = q v B sin 90 = q v B = const., since q, v, B are all const.

Direction of F is always perpendicular to v. Therefore the magnitude of v will not be changed by the force.

So we have a constant force acting on the particle, always perpendicular to its velocity, v.

These are conditions for circular motion – the particle will move in a circular path, whose spin axis is in the direction of B.

x

y

z

Bv

F

q


Radius of Circular Path, r:Using Newton 2: ΣF = ma

( )

B q vm

r

rv

m qvB

a mforce magnetic2

=

=

=

since path is circular

Radius, r, of the circular path is proportional to v (the particle velocity), inversely proportional to B (the strength of the field), and proportional to the ratio of mass to charge, (m/q).

This is an important relationship, in Physics, and has been used to measure the mass-to-charge ratio (which is a fundamental quantum constant) for the electron, and other elementary particles.( v, B, and r are measured, allowing the computation of (m/q).

y

z

v

F=qvBm

r

looking in the (–x)-direction


General Case:We have considered the case where the particle is injected into a uniform field at right angles to the field. What happens if the particle is injected at an angle θ to the field vector, B?

x

y

z

B

v

Fq

θ

Say B is in the +x-direction, and v is in the xz-plane, at angle θ to B.

Magnetic force, F, will be in the +y-direction, as before.

v can be replaced by its components v sinθand v cosθ , perpendicular to, and parallel to B, respectively.

v cosθv sinθ

The v sinθ component will produce the circular path, as before – the magnetic force, F, is due entirely to this component. The v cosθ component will simultaneously transport the particle, at a constant speed, in the +x direction.

The particle will consequently follow a helical path, spiraling about the +x-direction.


The Lorentz Force:We have considered the magnetic force on a charged particle in a magnetic field. If both magnetic and electric fields are present, both electric, and magnetic forces can act simultaneously on the particle. In this case, the overall electromagnetic force that acts is called the Lorentz force, and is the sum of the electric and magnetic forces.

F = q E + q v x B

where: F = the Lorentz forceq = charge of the particlev = velocity of the particleE = electric field vectorB = magnetic field vector

The controlling of the movement of charged particles, using the Lorentz force, finds some important instrumental applications. We will look at some of these now – in particular -

1. the velocity selector

2. the mass spectrometer

3. the cyclotron


1.Velocity Selector:

Hence, this device acts as a velocity filter by transmitting only those charged particles with a chosen velocity, and blocking all the rest.

We arrange the E and B fields such that the electric and magnetic forces are opposite in direction. For a particular charged particle, the electric force depends only on the charge, whereas, the magnetic force depends on both the charge and the velocity of the particle.

When produced, charged particles generally have a range of velocities.

A velocity selector is used to select from charged particles with a range of velocities, only those with a specific velocity.

It is based on crossed electric (E) and magnetic (B) fields.

Thus, the electric and magnetic forces will only balance each other, for particles having a specific velocity.

Our first application is the velocity selector.


The Crossed Fields:

N SB

Consider the uniform magnetic field between a N-pole and a S-pole.

Cross this field with the uniform electric field between two oppositely-charged, flat, parallel, metal plates.

E

Consider a particle, with charge q, moving between the pole faces, and the charged plates, with a velocity, v, that is perpendicular to both field directions.

+

v

q

q+B

E

FE

FB

vBecause of the directions of the fields, the electric force, FE, and the magnetic force, FB, will be in opposite directions, as shown.


Selection Condition:

q+B

E

FE

FB

v

If the electric and magnetic forces are equal in magnitude, a charged particle will pass undeflected

( ) ( )

( )1........... BE

v

B vq Eqforce Magnetic force Electric

=

==

If a particle’s velocity satisfies the selection condition, (1), it will pass through the field region undeflected; all other charged particles (with other velocities) will be deflected. Hence this system can be used to select charged particles with a chosen velocity, and act as a velocity filter. The undeflected particles pass through an aperture in a plate; deflected particles are blocked.

Note: that the velocity selected, v, in (1), is independent of the charge, q, and the mass, m, of the charged particles. It depends only on the ratio of the strengths of the electric and magnetic fields.


2.Mass Spectrometer:

The mass spectrometer is a widely-used scientific instrument, that allows for the analysis of the composition of a sample – it measures what atoms are present in the sample, and the percentage of each type of atom present. That is, it gives the distribution (or spectrum) of atoms present.This device uses electric and magnetic fields to deflect charged atoms (ions). It consists of three sections – an ionizer, a velocity selector, and an analyser. We look firstly at the analyser section.

B

Consider a region of uniform magnetic field, with magnitude B. Say we inject a charged particle, of mass, m, and charge, +q, into the field, perpendicularly to vector B.

v+

As described earlier, this particle will follow a circular trajectory, with radius, r, given by -

( ) ( )

.(2).......... Bv

qm

r

rv

m B vq

force lCentripeta force Magnetic2

��

��

�=→

=→

=

r

Our second application is the Mass Spectrometer.


The Analyser:...(2) Bv

qm

r ��

��

�=

B

v+

rIf we could fix the values of v and B, equation (2) shows that the radius, r, depends on the ratio of mass-to-charge (m/q). The reciprocal of this, the charge-to-mass ratio, (q/m) is traditionally used. By measuring r, and knowing v and B, we could compute (q/m), which is characteristic of a particular ion, and can therefore be used to identify the ion.

Fixing v :We need to fix the velocity, v, of an ion, independently of its charge and mass. This can be achieved using a the velocity selector, as discussed earlier, where the ions are directed through known, uniform, crossed electric, and magnetic fields. We can select the velocity of ions, v, according to the selection condition, derived earlier:

( )1........... BE

BE

vS

S==

Where ES and BS are the magnitudes of the crossed electric and magnetic fields in the Velocity Selector section of the mass spectrometer.


The System:

vIonizer

B

v+

r

Analyser

We now put the three components of the mass spectrometer together.

Firstly, we need to ionize the atoms we wish to analyse. This is typically done by passing an electric current through a vapor of the atoms, in a vacuum. This causes electrons to be knocked off the atoms, thereby ionizing them.

We then direct the ions through the perpendicularly crossed electric (ES) and magnetic (BS) fields of the Velocity Selector.

+

-VelocitySelector

ES

BS

The ions then enter the Analyser section, where they follow circular paths, whose radius, r, depends on the charge-to-mass ratio of the particular ion.

D

A linear detector, D, measures the numbers of ions, with various radii.

Thus the spectrum of ion (q/m) ratios, and hence atom masses can be measured.Note: the whole system must be under vacuum, so that air particles don’t scatter the ions under analysis, or contaminate the results.

AOnly ions with a selected velocity, will be undeflected, and be able to pass through aperture A.


(q/m) Ratio: ...(2) Bv

qm

r ��

��

�=( )1...........

BE

BE

vS

S==

B B

Eqm

rS

S��

��

�=(1) � (2) gives:

We set ES , BS and B. Then by measuring the values of r that occur, we can determine the corresponding (m/q) values, and so identify the ions present.

A variation of this instrument was used, in 1897, by J.J.Thompson, to measure for the first time, the charge-to-mass ratio for the electron.

where: r = radius of ion path in analyserEs = electric field in velocity selectorBs = magnetic field in velocity selectorB = magnetic field in analyser

and (q/m) = the charge-to-mass ratio of an ion

ES, BS and B are fixed, so r will be proportional to (m/q). Because only certain ions are present, with characteristic (m/q) values, only corresponding discrete values of r will occur.

By counting the number of ions at each radius, we can determine the ionic composition of the sample.


3.The Cyclotron:

This is a device used to accelerate charged particles to high velocities. The large kinetic energies of fast particles allows them to collide and interact at a fundamental level with other particles. Studies of such interactions allow us to investigate physics at a fundamental quantum level.The cyclotron uses electric and magnetic fields to accelerate charged particles in a vacuum.It consists of two hollow, metal D-shaped containers, with a gap between them.A uniform magnetic field, B, cuts down perpendicularly through the plane of the D’s.

An oscillating rectangular voltage, v, is connected across the D’s, so that an electric field exists in the gap between them. This field periodically is directed from D1 to D2, then reverses.

t

VT

V

Our third application is the cyclotron.

D1 D2

vacuum B

S

vT=period of the voltage


Path of a Particle:

t

V Vm

T/2 T 3T/2 2T 5T/2

Because they are static hollow metal containers, there will be no electric field inside the D’s, only between them.

E=0

Say a (+) particle is released near P, in the middle of the gap between the D’s, when D1 is (+) [ between t=0 and t=T/2]. The particle will be accelerated across the gap, and into D2.

When inside D2, because of the magnetic field (and zero electric field) the particle will follow a semi-circular path, that will bring it back to the gap.

We tune the oscillating voltage, V, so that D1 is negative, when the particle returns to the gap, and, for a second time, it will be accelerated across the gap, and into D1.

Inside D1, the particle will again follows a semi-circular path.

D1 D2

B

S

vmax

P

Q


Acceleration of the Particle:

t

vV

T/2 T 3T/2 2T 5T/2

Thus the particle will traverse the gap numerous times, being accelerated each time, by the electric field in the gap.

Each successive semi-circular path inside a D will have a greater radius, because of the particle’s increasing velocity.

Eventually, the radius of the particle’s path will increase to the outer radius of a D, and the particle will emerge from the cyclotron, at high velocity, vmax, (at Q).

For this to work, we need to get the correct period, T, for the oscillating voltage, v.

D1 D2

B

S

P

Qvmax


Period, T, of Oscillating Gap Voltage:

( ) ( )

( )′��

��

�=

��

��

�=→

=→

=

2.......... Br mq

vgTransposin

.(2).......... Bv

qm

r

rv

m B vq

force lCentripeta force Magnetic2

( ) ( )

( ) ( ) ( )4................ 2

qm

T gives 32

3.....v

r 2

velocitydistance

Tcircuit 1for Time

Bππππ

ππππ

��

��

�=→′

==≡

As we have seen before, the radius, r, of the semi-circular path of a charged particle in a uniform magnetic field ( such as inside a D ) is given by -

where: q = particle chargem = particle massB = field in the Dv = velocity of the particle

To ensure that the polarity is correct for an acceleration of the particle in the gap, the period, T, of the oscillator needs to be synchronised with the time for a complete circuit of the two D’s.

Particle in a D


KE of Particles Emitted by Cyclotron:

( )4................ 2

qm

TBππππ

��

��

�=

( )

( )

( )5........ m 2

B R qKE

2 from B Rmq

m21

vm21

particle emitted of KE

222

max

222

2max

=→

′��

��

�=

=

For a given particle (given (q/m) ratio), with the magnetic field set to a value, B, the period of the oscillating voltage must be tuned to satisfy (4).

Note: T is specific to a particular particle, but is the same for all circuits of that particle within the D’s ( whatever the v, and r).

t

vV

T/2 T 3T/2 2T 5T/2

vmax = particle’s exit velocity

where R = largest radius within a D= r corresponding to vmax

To maximise KE, we need large B and R.

( )′��

��

�= 2....... Br mq

v

The KE of emitted particles will be determined by their exit velocity from the system.


Example:

MHz 26 Hz 106.2T1

f

s 108.37.1

2106.11067.1

B2�

qm

T

7

819

27

=×=≡∴

×=��

��

�

××=��

�

��

�= −

−

− ππππ

A small cyclotron has D’s of radius 0.25 m, and is used to accelerate protons. The magnetic field inside the D’s is 1.7 T. Find the oscillator frequency required between the D’s, and the KE of protons emitted.

Oscillator Frequency: Given: R = 0.25 m B = 1.7 T

From (4):

For a proton, q = 1.6x10-19 C & m = 1.67x10-27 kg

Proton KE:

( ) ( ) ( )( ) J 104.1

101.67 20.25 1.7 106.1

2mR B q

KE 1227-

22219222

max−

−

×=×

×==From (5):


Magnetic Force on Currents:So far, we have considered the magnetic force on a freely moving charged particle. In practice, charges often move through magnetic fields, as part of a current, which is constrained to flow in a conductor. In this case, the magnetic forces are transferred to the conductor. This is the principle of operation of the electric motor.

Clearly, the total magnetic force on the conductor, will be the sum of all the component magnetic forces, on all the charges, that make up the current flow. Since these component charges may be traveling in different directions, in different parts of the conductor, relative to the magnetic field, the total force will depend on the detailed shape of the conductor, and the geometry of the magnetic field.

To simplify this complex situation, we consider a tiny displacement, at a given point along the conductor called a “length element”, ds, which, because of its infinitesimal length, must be straight. The direction of the ds vector is taken to be the direction of current flow, thus representing the direction of the velocity of the charges, moving in the element.

To find the total magnetic force on a real length of wire carrying a current in a magnetic field (such as in an electric motor), we need to do a vector addition of the forces on all the length elements that make up the conductor.

ds

dsii

v


Force on Displacement Element, ds:

( ) ( ) ( )BdsBdsBdsBvF ×=×=��

��

� ×=×= i dtdq

dt

dq dqd

Consider a displacement element, ds, of the conductor. Say ds points in the +y-direction

dsB

dF

x

y

zP

Q

i

Say a conductor, carries a current, i, from P to Q, in a magnetic field, as below.

Say, at ds, the B-vector of the field points in the +x-direction.Then (from F = q v x B) the total magnetic force, dF, on all the moving charges within the element, will act in the (–z)-direction.

At any moment, there will be some total charge dq flowing in the length element, ds, at an effective velocity v. Let dt be the time interval for all the charge, dq, to travel through the length element, ds.

( )BsF ×= d i d Magnetic force on current element (cross product).

( Of course, ds could point in any direction, relative to B.)


Force on a Length:

dsB

dF

x

y

zP

Q

i

Say a conductor carries a current, i, through a region of magnetic field, from P to Q.To get the total magnetic force, F, on the total length of conductor, we need to do a vector addition of all the elementary forces, dF, on all the length elements, ds, that make up the conductor from P to Q.

( ) ( )�� ×==≡Q

P

Q

P

id force Total BdsFFi = const. for all elements, over the summation

This is called a “line integral”, since we add up the elementary forces, dF along the conductor. Depending on the direction of the particular ds, and local B, the direction of the dF’s will, in general, vary from point to point along the conductor.


Straight Length in Uniform Field:

B

dF

x

y

z

P

Q

ids

Say the length of wire in the field, PQ, is straight, and of displacement L. (L is the sum of all the ds’sfrom P to Q).

( ) ( )

( )

( ) ��

��

�=×=→

=×�

��

=

×==≡

�

�

��

LdsBL iF

constB since Bds

BdsFF

Q

P

Q

P

Q

P

Q

P

since

i

id force Total

Say the field between P and Q is uniform (B = const).

L = vector PQ.(L is in the direction of current i).

(Because all ds’s are parallel, and all B’s are parallel, all the dF’s are in the same direction, which consequently is the direction of their sum, F.)

Magnitude of F = i L B sin θ , and is in a direction given by the right-hand-rule for cross products

L


Closed Loop in Uniform Field:

B

dF x

y

z

i ds

Here we work out the total force, F, on a closed loop, that carries a loop current, i, in a uniform magnetic field.( the loop is entirely within the field, but can have any shape ).

( )

( )( ) loop closed a round zero is since 0

field uniformfor since

i i

loop around forces elementary of sum

loopon force Total

�

��

�

=

��

��

� =×=×=

��

��

�=≡

ds

constantBBdsBds

dFFThe total magnetic force on a closed, current-carrying loop in a uniform magnetic field is zero. However, the total magnetic torque on the loop is not zero.

i

B

dF

dFNote: the elementary forces will be in opposite directions, on opposite sides of a symmetrical loop.


Rectangular Loop in a Uniform Field:

x

y

z

P

S

R

QBθθθθ θθθθ

We consider such a loop, with sides of lengths a and b, and area A = ab, that carries a loop current of i.

A uniform magnetic field, B, passes through the loop at angle θ to area vector A.

(A RHR convention works for the direction of A. If the fingers of the right hand curl about the loop in the direction of the loop current, the thumb points in the direction of area vector A.)

If we take displacement vectors along sides of the loop, in the directions of current flow, then we can define –

a = PQ & b = QR

Vector area of the loop = A = a x b (cross product)

i

a

b

A

i

a

b

A

i

i

The total magnetic torque on a rectangular loop in a uniform magnetic field is the basis for the electric motor, and therefore has particular practical importance.


Forces on the Loop:

F

(-F)

(-f)

fx

y

z

P

Q

R

SBθθθθ θθθθ

i

a

b

A

Hence, the magnitude and direction of the force on each side of the loop is given by -

F = i L x B

(The RHR gives the directions of these forces –F is in the (–y)-direction; f is in the +z-direction. )

( ) ( )( )2.....i1.....i

Bb f Ba F-

×=×=

We have previously shown that the total force on any closed loopcurrent is zero. This can be confirmed from the Figure, since –

(Total force) = +F –F +f -f = 0

The total magnetic torque on the loop, however, is not zero. We will now find this torque.

The magnetic force on a straight current-carrying conductor, in a uniform field is given by -,


Torque on the Loop:We take torques about the symmetrical centre point of the loop, C. We use the definition of torque, ττττ, namely τ τ τ τ = r x Fwhere r is the perpendicular displacement from C to the side of the loop on which the magnetic force acts.

(Total torque due to f, -f, F, -F) = ττττ=( ½ a x f) + (( - ½ a ) x ( -f)) + (½ b x F) + (( - ½ b ) x ( -F))= 2 (½ a x f) + 2 (½ b x F)= 0 + (b x F)

Therefore: ττττ = (b x F) ....................(3)

since a is parallel to f, & therefore |a x f| = a f sin 0 = 0

( f & ( -f ) try to stretch the loop in the direction of the z-axis, but exert no torque upon it; F & (-F) exert all the torque.)

F

(-F)

(-f)

fx

y

z

P

Q

R

SBθθθθ θθθθ

i

a

b

A

½ b

a

b

i

F

(-F)

(-f)

f½ a

C


Total Torque on Loop:

τ τ τ τ = (b x F) .....(3)

View the loop, edge-on, looking in the (–z)-direction

Magnitude of ττττ:

Direction of ττττ: From the RHR, (b x F) is in the (-z)-direction.

Both magnitude, and direction of ττττ are consistent with the following expression for the magnetic torque on the rectangular loop.

ττττ = i (A x B) (cross product)

F = i a x B .....(1)

F

(-F)

(-f)

fx

y

z

P

Q

R

SBθθθθ θθθθ

i

a

b

AA

F

Bb θ

θ x

yS

Rθ

|τ τ τ τ | = τ = |b x F| = b F sin θ [since angle from R to F is θ ...see figure].= b ( i a B ) sin θ from (1), since a is perpendicular to B]

= i (a b) Β sin θ = i Α Β sin θ [since A = ab]


Generalisation :

ττττ = i (A x B) (cross product)

We have derived this result for the magnetic torque, ττττ, on a rectangular loop, of vector area, A, in a uniform magnetic field, of vector B. However, it is true for a flat loop of any shape, that carries a current, i, in a uniform field.

If the loop is replaced by a coil with N turns, then the torque will be given by –

ττττ = N i (A x B)

This formula is the basis for the electric motor, and also of the moving-coil galvanometer, from which the moving-coil ammeter and voltmeter are derived.

(The torques for individual loops of the coil are additive.)


Example:

x

y

z

P

Q

R

S

I

a

b

A

Bθθθθ θθθθ

4m

3m

2m

The loop PQRS, shown, carries a current I = 10 A, in a uniform magnetic field of 2 T. The field is directed in the (+x)-direction. Find the magnetic torque, τ, on the loop.

B = 2 i (T)

Area vector of the loop = A = (A cos θ ) i + (A sin θ ) j=(ab cos θ ) i + (ab sin θ ) j

Torque = ττττ = I (A x B) = I [(ab cos θ ) i + (ab sin θ ) j ] x [ 2 i ]= I (ab sin θ) (2) (-k) = 10 (2 x 5 x 4/5) (2) (-k) = - 160 k (N m)

Torque is 160 N m, directed in the (-z)-direction.

( If the loop starts from rest, the z-axis will be the spin axis of the loop.)


i � 0 Σ=•� dsB


What Creates a Magnetic Field?:In our discussion of the magnetic field, we firstly defined the field in terms of a vector, B. We then looked in detail at the properties, and applications, of the magnetic force. We now ask the fundamental question - what created the magnetic field in the first place?.

At a fundamental physical level, a magnetic field is observed, when a charge moves, relative to an observer. If the charge is stationary, only an electric field is detected, when the charge moves, an additional magnetic field is detected, by the observer. The field is due to the relative velocity between a charge, and an observer.

Magnetic fields are produced by electric currents.

But moving charges constitute a current - so

Because a current flows along a line in space (that is, the current is essentially one dimensional), the field produced by the current must have axial symmetry –it must be symmetrical about the line. So magnetic fields curl about the current that produces them.


Direction of the Magnetic Field due to a Current:Consider a long straight current, i.

i

The magnetic field (flux lines) curl around the current direction.

eye

From the eye’s view -

The magnetic flux lines are circles, coaxial with the current.The direction of the flux lines is given by the Right Hand (Grip) Rule (RHR.)

Clasp the current in the Right Hand, with the thumb pointing in the direction of current flow. The fingers will then curl about the current direction in the direction of the magnetic flux.

eye’s view

current, i, upward

i


Biot-Savart Law - Setup:

P

Say a conductor carries a current i

We want to find the elemental magnetic field, dB, at a point P, produced by the current-carrying element ds, of the whole conductor.

In 1820 J.B. Biot & F. Savart measured the experimental relationship between a current and its magnetic field - the Biot-Savart Law.

Using the RHR for magnetic field direction the dB-vector will be downward, at P.

r

dBSay that the position vector of point P from the displacement element, ds, is r .

Because the field produced by a current-carrying conductor, at a point, P, depends both on the length, and on the shape, of the conductor, we consider a short, straight displacement element, vector ds, of the conductor. ( ds is taken in the direction of current flow.)

Vector ds can be resolved into components (ds cos θ), in the direction of r (ie. the direction of P), and (ds sin θ), perpendicular to rExperiment shows, that the component in the direction of P, (ds cos θ), contributes zero field at P; dB is due entirely to the perpendicular component, (ds sin θ).

ds sinθθθθ

ds cosθθθθ θ

ids


Biot-Savart Law :

( )�� ˆproduct cross theof that asdirection same the

be out to turnsand RHR, by thegiven is ofdirection The

×dB

i

P

ds

r

dB

where: (µ0 /4π) is the constant of proportionality for the SI unit system.µ0 is a magnetic constant called the “permeability of vacuum”.

(µ0 = 4 π x 10-7 N A-2 :exact SI value )

θBiot & Savart found that the magnitude of dBis proportional to -

- the current i

- the (ds sin θ) component of the element ds

- the inverse square of the magnitude of r

20

2

r

sin� i

�4�

r

sin� i

ds

dsdB

=

∝So, the magnitude of dB can be expressed by -

θ is the angle from ds to r


Vector Form :

20

rˆ

i �4� rdsdB ×=

. vector ofdirection in ther unit vecto theis ˆ and rr

i

P

ds

r

dB

Taking account of the experimental facts of the previous slide, both magnitude and direction of dB, at P, can be incorporated into a single vector expression, which is the general form for the Biot-Savart Law:

crossproduct

where: µ0 is a magnetic constant called the “permeability of vacuum”.(µ0 = 4 π x 10-7 N A-2 :exact SI value )

θ

The Biot-Savart law plays an analogous role in magnetic fields, to that played by Coulomb’s law in electric fields – both give the link between the field, and the source of the field.


Biot-Savart Law For an Entire Current :

��×==

pathcurrententire

20

pathcurrententire r

ˆ i

�4�

rdsdBB

i

P

ds

r

dB

Biot-Savart gives the field due to one element, ds, of a current path. To find the total field, B, due to the entire current path, we need to sum all the field contributions, dB, from all of the elements, ds, along the path.

This integral is a vector summation of all the dB’s produced at a given point P, by all of the segments of a line (a “line integral” along the current path).

20

rˆ

i �4� rdsdB ×=

We will now apply the Biot-Savart law to find the magnetic field produced by two simple current geometries.

(since i is constant for all path elements)


Current Loop:

( )1...... Rds

i �4�

r

ˆ i

�4�

dB 20

20 =

×=≡

rdsdB

20

rˆ

i �4� rdsdB ×=

As a first application of the Biot-Savart law, we will find the magnetic field at the centre of a flat, circular, current loop that carries a constant currant.

iR

Consider a circular current loop, of radius R, that carries a constant current i.

ds

dB

Let ds be a typical displacement element of the loop.

Let dB be the element of field produced at the centre of the loop by the current in ds.

By symmetry, and the RHR, dB must be perpendicular to the plane of the loop, as shown. This direction is consistent with the direction of the cross product (ds x r).

The full dB vector is given by the Biot-Savart law:

Here, ds is perpendicular to r, and |r| = const = R. Therefore the magnitude of dB = |dB| is given by:

ds 90sin 1 ds

ˆo =××=

×rds


Total B at Centre: ( )1...... Rds

i �4�

dB 20=

( )

R 2i �

loopcircular of ncecircumfere

R 2 R �4i �

ds R �4i �

dB B

0

20

loopcurrent circular around

20

pathcurrententire

=

��

��

←=

== ��

ππππ

iR

ds

dB

The total magnetic field vector, B, at the centre, is the vector sum of all the individual dB’s, due to each of the length elements, ds, composing the current loop.

B

But, for this case, by symmetry, and by Biot-Savart, all the individual dB’s will be in the same direction. Therefore we can add the individual magnitudes, dB, to get the total magnitude B, knowing that the vector B will have the same direction as any of the individual dB vectors.

The total field, B, at the centre of the loop, has magnitude

R 2i �

B 0=

and is perpendicular to the plane of the loop according to the RHR.

Summary:


Long Straight Current - Setup:

( )1..... rR

�)-sin(� sin�

=

=

As a second application of the Biot-Savart law, we will find the total magnetic field vector, B, at a radial distance of R, from a long, straight, constant current of i.

X

Y

P

rR

x

θ

dB

dx

i

( )( )2.....

xR

�-�-tan �tan

−=

=

Consider a displacement element, dx, of the current, at coordinate x.Let P be a point on the y-axis, at a radial distance of R from the current.Take the x-axis along the current direction.

Let r be the displacement vector from dx to P.θ is the angle from the (+x)-direction to the direction of vector r.Let the element of magnetic field produced by dx, at point P, be vector dB.From the diagram -


Find dB:

X

YP

rR

x

θ

dB

dx

i

( )1..... rR

�sin =

( )2..... xR

�tan −=

By the RHR, the direction of dB is perpendicularly upward from the plane of the figure.

The magnitude of field element dB is given by the Biot-Savart law -

( )3......dx r

sin�

4�i �

rsin� 1dx

4�

i �

r

ˆ

4�i �

dB

20

20

20

=

××=

×==

rdxdB

To find the total field, B, at point P, we firstly find the element of field, dB, produced at P, by the current in the length element dx

Equation (3) gives dB in terms of three variables, θ, r, and x. But θ, r, and x are dependent variables – as one of them varies, the other two vary, as functions of the first.

We need to express two of θ, r, and x, as functions of the other variable.

It is easiest to change the variable in (3) to θ.


Find dB in terms of θθθθ:

( )3......dx r

sin�

4�i �

dB 20=

( )4..... �sin

Rr =

......(5) d� �sin

R dx

�sin

R

�cos1

�sin�cos R

�sec �tan

R

d�dx

�tan

R x

2

2

22

2

22

=∴

=

=

=∴

−=

( )1..... rR

�sin =

( )2..... xR

�tan −=

Find dx as a function of θθθθ:

Find r as a function of θθθθ:

From (1):

From (2):

(4),(5) --> (3):

6).........( d� sin� R 4�i �

d� �sin

R

R�sin

1

sin�

4�i �

dB

0

22

20

=

=

This gives the magnitude of the field element produced, at P, by the displacement element, ds, of the current path, for which the bearing of point P is θ.


Find B:6).........( d� sin�

R 4�i �

dB 0=

[ ]

[ ]

( ) ( )[ ] ......(7) R �2i �

1-1 R �4i �

0cos-cos� R �4i �

onst.R �4i �

since cos�- R �4i �

d� sin� R �4i �

dB B

00

0

0��

0�0

��

0�

0x

-x

=+−−=

−=

==

==

==

=

=

+∞=

∞=��

c

X

YP

rR

x

θ

dB

dx

i

To compute the total field, B, at point P, we need to add all the elementary field components dB, from all segments dx along the total length of the current:All dB ‘s point in the same direction, and therefore their magnitudes can be added, to get the total B.

When the dx is at the left end of the line of current, θ = 0, when at the right end, θ = π

( )7..... R �2i �

B 0=


Summary:

R �2i �

B 0=

YP

R

dB

i

The magnetic field produced by a long, straight, constant current, curls symmetrically about the current direction.

Its magnitude, at any point, is proportional to the magnitude of the current, i, and inversely proportional to the radial distance, R, from the current.

Its direction at any point is given by the RHR.

B

R

i = const


Ampere’s Law:

Biot-Savart gives the magnetic field vector, in terms of the current that produces it. It is the magnetic analogue to Coulomb’s law, which gives the electric field vector, in terms of the electric charge producing it.

However, Biot-Savart, like Coulomb, becomes mathematically difficult when applied to many practical cases.

Ampere’s law is a powerful reformulation of Biot-Savart, that allows many otherwise difficult cases, to be easily solved. Ampere’s law is the magnetic analogue to Gauss’ law for electric fields. It exploits the essential symmetry present in many situations, to facilitate their solution.

We will derive Ampere’s law for the particular case of the field due to a long straight current. Our result, however, has general applicability.

We start, by making an observation about our result for the long, straight current.


Observation:

ieye

Consider the magnetic flux lines about a long straight current.

These flux lines are circular. Consider one with radius R.eye’s view

R

( )( ) i �R �2 B

R �2i �

B : 7

0

0

=→

=where (2πR) is the length around the circumference of a circular flux line, and B is the (const.) magnitude of the field vector, along that flux line.

Our previous derivation, for this case, equation (7), gave

Conclusion: Multiplying B by the displacement around a flux line, is proportional to the current enclosed by the flux line.


More Generally:

eye’s view

R

( )

i �

R 2 R 2i

ds B

B.ds

0

0

=

��

��

�=

=

=

•

�

�

�

ππππππππ

µµµµ

dsB

ds

Lets generalise our observation, by summing the dot product B.dsaround the flux line (that is, the line integral of B around the flux line).

( )7.... R 2i

0

ππππµµµµ=B

Let ds be an element of path around the flux line. Let B be the field vector at that element of path.

B

( )path closed a around

points, allfor of sum a is B.ds�

since B is parallel to ds at all points around the flux line, B.ds =|B|.|ds|.

since |B| = const. at all points on the flux line, by symmetry.

sum of all the ds segments is the circumference of the circular flux line =(2 π R).

From (7).


Generalization:i � 0=•� dsBFor a long, straight current, we have shown that

The line integral of (B.ds) around a flux line is proportional to the current enclosed by that flux line (that is, the current that produces the flux line and the associated field vector B).

This turns out to be valid for any closed path that encloses any constant currents.

µ0 is the permeability constant = 4π x 10-7 N A-2

( )8.... i � 0 Σ=•� dsB

The line integral of B.ds around any closed path is equal to µ0 Σ i , where Σ i is the algebraic sum of all the currents that cut through a surface that is bounded by the closed path, and µ0 is the permeability constant

Such a closed path is called an Amperian Path (AP). The AP is composed of displacements elements, ds, of path, and at a particular element ds, the field vector is B.

In General:Ampere’s Law

Equation (8) is a general statement of Ampere’s law.


Notes:ΣΣΣΣ i : Σ i is the algebraic sum of the currents that cut a

surface bounded by the AP (an Amperian Surface).

I IAP

Σ i = I + I = 2I

The positive current direction is taken to be consistent with the RHR. If we curl the fingers around the AP in the direction of travel for the line integral, then the thumb will point in the positive current direction.

I

I

AP

Σ i = I - I = 0

Note: that Ampere’s law assumes all currents to be constant.

2A

5A

AP

Σ i = +2 - 5 = -3A


Notes:

AP: In principle, an AP of any shape can be used,. However, to facilitate a solution, it is important to choose a shape such that the evaluation of B.ds is as simple as possible, for all points around the AP.

The evaluation of B.ds will be simple if

(i) B is parallel to ds (for example, if we follow a flux line)(ii) B is perpendicular to ds ( if we cut perpendicularly, across flux lines )

(iii) |B| = 0 (the magnitude of B is zero, or the field is very weak)

To achieve these simplifying conditions, we need to choose an AP whose shape corresponds to the symmetry of the field.

Steps in using Ampere’s law:1. Draw a diagram, showing the magnetic flux lines.2. Choose an optimally shaped AP.3. Write down Ampere’s law.4. Integrate Ampere’s law around the AP, carefully justifying all steps, to find B in terms of currents, i.


The Long Straight Current Revisited:As our first application of Amperes law, we will recompute our expression for the magnetic field vector, B, at a radial distance of R, from a long, straight, constant current, i.

ieye

eye’s view

R

Draw the AP to follow the flux line of radius R.The AP is a circle of radius R, coaxial with the current.

AP

We will integrate around the AP in the direction of the flux.

ds

B

We show a typical displacement element, ds, of the AP, and the corresponding B vector.

B will be parallel to ds at all points around the AP, since we are following a flux line.

Also, around the AP, the magnitude of B will be constant, even though its direction changes. This follows from symmetry, since at all points around the AP, the current has the same relative orientation, and is the same distance away.


Find B:

eye’s view

R

AP

ds

Bi

( )

R 2i

B

i R 2 B

i ds B

i B.ds

i

0

0

0

0

0

ππππµµµµ

µµµµππππ

µµµµ

µµµµ

µµµµ

=

=

=

=

Σ=•

�

�

� dsB Ampere’s law

since B parallel to ds along a flux line,and thus B.ds = |B|.|ds|

i is the total current.

since |B| = const., by symmetry.

AP is a circle, of radius R, and the sum of all scalar segments is the circumference = (2πR).

This is the same result that we derived earlier, using the Biot-Savart law, but the derivation here, has been much simpler.

We now invoke Ampere’s law.


Field Inside a Solenoid:A solenoid is a coil of wire, wound on a cylindrical former. For our second application of Ampere’s law we will derive an expression for the magnetic field inside such a solenoid. This derivation would be difficult using Biot-Savart, but is greatly simplified by using Ampere.

Say the coil has a total of N turns, uniformly wound over a total coil length of L.

ii

L

N turns

Let the number of turns per unit length = n = N/L

Say the current in the solenoid is i.

The flux loops of the magnetic field pass through the centre of the solenoid, where they are confined, and are close together, producing a strong field within.

Outside of the solenoid, the flux loops can spread out, resulting in a weaker field outside.

Consider the solenoid’s magnetic field.


Direction of the Internal Field:

ii

L

N turns

The direction of the field within the solenoid is given by a version of the Right Hand Rule (RHR)

This rule for the direction of the internal magnetic field of a solenoid, is a consequence of the RHR for a single conductor. The fields for the individual adjacent conducting turns of the coil, merge together into the overall solenoid field.

Hold the solenoid so that the fingers of the Right hand, curl about the solenoid in the direction of the current in the turns. The thumb will then point in the direction of the internal field through the solenoid.

By symmetry, the flux lines within the solenoid will be parallel to each other.


The AP:

ii

L N turns

For the AP we choose a rectangle, of length d, inserted into the solenoid as shown – the top side is inside the solenoid; the bottom side is outside.

d

Let the field vector along the top side of the rectangle be B.

d

dsds

ds

ds

BBB

B=0

Integrate around the rectangle in the clockwise direction.

Directions of B at each side are shown.

The bottom side of the rectangle is outside the solenoid, where the field is weak (B = 0)

B


Find B:

d

dsds

ds

ds

BBB

B=0

Invoke Ampere’s law.

( )

( )in B

i nd d B

i nd ds B

i 0 0 0 B.ds

i

i law sAmpere'

0

0

0top

0top

0

0

µµµµµµµµ

µµµµ

µµµµ

µµµµ

µµµµ

==

=

Σ=+++

Σ=•+•+•+•

Σ=•

�

�

��

�

L.sidebottomR.sidetop

dsBdsBdsBdsB

dsB

B perpendicular to dsalong L. and R. sides

B = 0 alongbottom

since B || dsalong top.

Integrating along each side of the rectangle

|B| = const. along top by symmetry.

there are (nd) turns cutting the Amperianplane, each carrying current i.

top side length = d


Conclusions:

in �B 0=The magnitude, B, of the internal field inside a solenoid, is proportional to the current, i, in the windings, and to the number of turns per unit length, n.

Because B is independent of the location of the top side of the rectangular AP, within the solenoid, the internal field must be uniform.

Note: B is independent of the diameter of the solenoid.

Example: Find the internal field of a solenoid, that has 500 turns wound over a length of 10 cm, when it carries a current of 2 A.

B = µ0 n i = (4π x 10−7) x (500/0.1) x 2 = 0.126 (T)


Conclusions:In our considerations of the magnetic field, we firstly quantified the field, by defining a magnetic field vector, B, using the magnetic force on a moving charge.

We then considered various situations, and applications, involving the magnetic force, on both moving charges, and current-carrying conductors.

Finally, we looked at current as the source of the magnetic field, and the law connecting the current to the field.


N S

4/4/2006 magnetic fields (© f.robilliard) 1flai/theory/lectures/magneticfields.pdf · 4/4/2006...

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