Sources of magnetic field

The magnetic field of a single moving charge

Our textbook starts by stating “experiments show that …”-That is a valid point, always the supreme criterion, and historically what happened.However, we have shown already in our relativistic consideration that

moving charges of a current create a B-field. Recall:

In the lab frame (frame of the wire) we interpret this as the magnetic Lorentz force

202vF qv qvB

R c

with 0 02

02 2 2v v IB

R c R R

magnetic B-field in distance r from a wire carrying the current I

2

0 0

1c

Next we show:the field of the straight wire can be understood in terms of the sum of the contributions from individual moving chargesThe field of a moving charge measured in P at r from the moving chargereads

034

qv rBr

For an infinitesimal small charge dq we get

034

dqv rdBr

0

34

dldq rdtr

0 0

3 34 4

dq dl r I dl rdtr r

dl

R

r 03

sin ( )4I r lB dB dl

r

02 2 3/24 ( )

I R dlR l

02 2 3/24 ( )

I R dlR l

0 02

4 2I I

R R

P

As we have shown before in the relativistic consideration

Auxiliary consideration (keep practicing):

2 2 3/ 2( )R dl

R l

22 3/ 2

1 1(1 / )

dlR l R

With substitution sinh x= l/R coshdl R x dx

2 2 3/ 2

1 cosh(1 sinh )R x dx

R x

2 2 2 3/ 2 2

1 cosh 1(cosh sinh sinh ) cosh

x dxdx dxR x x x R x

http://en.wikipedia.org/wiki/File:Sinh_cosh_tanh.svg

With2 2

2 2

sinh cosh sinh 1tanhcosh cosh cosh

d d x x xxdx dx x x x

2

1 1 2tanhcoshdx dx x

R x R R

Biot-Savart law

The vector magnetic field expression for the infinitesimal current element

034

I d l rd Br

is known as law of Biot and SavartIt can be used to find the B-field at any point in space by and arbitrary current in a circuit

034

I d l rBr

Biot and Savart law

Let’s consider an important example

Magnetic field of a circular current loop at a distance x from the center

034

I d l rd Br

How do we know this is Due to

dB r

Or you calculate for this particular element:

zd l dl e sin , cos ,0r r

dl r

0 0 cos sinsin cos 0

x y z

x y

e e edl r dl e r dl e

r r

We take advantage of the symmetry to conclude:The y-components of the B-field cancel out 0

3

cos4xI dl rdB

r

02

cos4xI dldB

r

with 2 2 2r a x and / cosa r

0

3/ 22 24xI dl adB

x a

2 20 0

3/ 2 3/ 22 2 2 2

24 2x xI a I aB dB

x a x a

For a thin coil of N loops we get for the magnetic field on the coil axis

2

03/ 22 22x

I N aBx a

and specifically at the center x=0

0

2xI NB

a

We can also express the field in terms of the recently introduced magnetic moment

2

03/ 22 22x

N I aBx a

Magnetic moment of N loops

See http://www.pnas.org/content/105/37/13716.full.pdf+html for complete article

Application of Bio-Savat law for potential future fusion technology

https://www.iter.org/mach

ITER (International Thermonuclear Experimental Reactor)based on tokamak concept

HSX (Helically Symmetric eXperiment)@Univ. of Wisconsin-MadisonModular coil stellarator

Video from February 2014 on ITER

Latest News April 15, 2015

Diagram of the QAS2 stellarator

Garabedian P R PNAS 2008;105:13716-13719

color map of the plasma surface

12 coils produce a magnetic field designed to confine the plasma in equilibrium.

Four of 12 modular coils that produce the magnetic field of the QAS2 stellarator using the Biot–Savart law.

Ampere’s law

Similar to the simplified calculation of electric fields of charge distributions using Gauss’ law we can often find a simplified way to calculate magnetic fields of currentsBiot-Savart works always but integration can be tough

In electric case we could use Gauss’ law

Magnetic Gauss law

0Bd A not so useful

Let’s explore what a line integral can do

0

QEd A

0Edr In the electric case because E conservative

What about

?Bdr

Let’s consider magnetic field caused by a long, straight conductor

0

2 tIB er

We chose as integration path the path of the field line

0 0 0

00

2 2 222

t t tI I IBd r e d r e e dr drr r rI r Ir

Line integral just depends on the enclosed current (not on r… )

What if the integration path does not enclose the current ?

We consider this example (result holds in general see textbook)

Bdr Bdr Bd r Bd r Bd r a b b

cc d d a

Bdr Bd r a b c d

00 0

1 21 202 2I Ird r dr r

0 0 02 2I I

Last step of generalizationWhat a positive or a negative current is with respect to the integration path is determined by the right hand rule

Curl fingers of right hand around integration path

Thumb defines direction of positive currentArbitrary closed path around conductors

Ampere’s law0 enclBdr I

Let’s apply Ampere’s law (we use the B-field version for now)

Of course we could use it to determine the B-field of a long straight wire

0Bdr I

02Bdr rB I 0

2IBr

Remark: Often you find in the literature Ampere’s law written in term of the H-field enclHdr I

In vacuum the relation between B and H is simple 0B HWhen the fields in matter are considered situation more complex 0B H M

magnetization

Radial symmetry of B-field is input

B-field inside a long straight wire

22 2

0 0 02 22 I IrBd r rB j r rR R

0 22IrBR

B-field of a long solenoid

Field is homogeneousinside a central part

Let’s have a look to a centralpart of the solenoid

Bd r Bd r a b

0BL NI

with n=N/L number of turn per unit length 0B nI

What can you say about the relation between the direction of the B-field in the solenoid and the direction of the magnetic dipole moment?

Clicker question

1) There is no relation

2) They point in the same direction

3) They are antiparallel

4) They are orthogonal

5) They make angle of cos-1B