Magnetic Fields Due to Magnetic Fields Due to Currents Currents

Chapter 29Chapter 29

Remember the wire?

Try to remember…

30

20 4

1

4

1

r

dq

r

dq

rd

rrE

VECTOR UNITr

r

The “Coulomb’s Law” of Magnetism

A Vector Equation

For the Magnetic Field,current “elements” create the field.

TmATm

typermeabilir

id

r

id unit

770

30

20

1026.1/104

44

:field fashion tosimilar aIn

rsrs

B

E

This is the Law ofBiot-Savart

Magnetic Field of a Straight Wire

We intimated via magnets that the Magnetic field associated with a straight wire seemed to vary with 1/d.

We can now PROVE this!

From the Past

Using Magnets

Right-hand rule: Grasp the element in your right hand with your extended thumb pointing in the direction of the current. Your fingers will then naturally curl around in the direction of the magnetic field lines due to that element.

Let’s Calculate the FIELD

Note:

For ALL current elementsin the wire:

ds X r

is into the page

The Details

02

0

20

)sin(

2B

it. DOUBLE and to0 from integrate

wesoamount equalan scontribute

wire theofportion Negative

)sin(

4

r

dsi

r

idsdB

Moving right along

R

i

Rs

RdsiB

SoRs

R

Rsr

22

)sin(sin

0

02/322

0

22

22

1/d

Center of a Circular Arc of a Wire carrying current

More arc…

Cpoint at 4

44

44

0

0

0

02

0

20

20

R

iB

dR

i

R

iRddBB

R

iRd

R

idsdB

Rdds ds

The overall field from a circular current loop

Iron

Howya Do Dat??

0rsd

0rsd

No Field at C

Cpoint at 4

)2/(0

R

iB

Force Between Two Current Carrying Straight Parallel Conductors

Wire “a” createsa field at wire “b”

Current in wire “b” sees aforce because it is movingin the magnetic field of “a”.

The Calculation

d

iLi

iFd

iB

ba

b

a

2F

angles...right at are and Since

2

:calculatedjust what weis a"" wire

todue b"" at wire FIELD The

0

b""on

0b""at

BL

BL

Invisible Summary

Biot-Savart Law (Field produced by wires)Centre of a wire loop radius RCentre of a tight Wire Coil with N turnsDistance a from long straight wire

Force between two wires

a

II

l

F

2

210

a

IB

2

0

R

IB

20

R

NIB

20

20 ˆ

4 r

rdsB

id

Ampere’s Law The return of Gauss

Remember GAUSS’S LAW??

0enclosedq

d AESurfaceIntegral

Gauss’s Law Made calculations easier than

integration over a charge distribution.

Applied to situations of HIGH SYMMETRY.

Gaussian SURFACE had to be defined which was consistent with the geometry.

AMPERE’S Law is compared to Gauss’ Law for Magnetism!

AMPERE’S LAWby SUPERPOSITION:

We will do a LINE INTEGRATIONAround a closed path or LOOP.

Ampere’s Law

enclosedid 0 sB

USE THE RIGHT HAND RULE IN THESE CALCULATIONS

The Right Hand Rule .. AGAIN

Another Right Hand Rule

COMPARE

enclosedid 0 sB

0enclosedq

d AE

Line Integral

Surface Integral

Simple Example

Field Around a Long Straight Wire

enclosedid 0 sB

r

iB

irB

2

2

0

0

Field INSIDE a WireCarrying UNIFORM Current

The Calculation

rR

iB

andR

rii

irBdsBd

enclosed

enclosed

20

2

2

0

2

2

sB

R r

B

R

i

2

0

Procedure

Apply Ampere’s law only to highly symmetrical situations.

Superposition works. Two wires can be treated separately

and the results added (VECTORIALLY!) The individual parts of the calculation

can be handled (usually) without the use of vector calculations because of the symmetry.

THIS IS SORT OF LIKE GAUSS’s LAW

#79 The figure below shows a cross section of an infinite conducting sheet carrying a current per unit x-length of l; the current emerges perpendicularly out of the page. (a) Use the Biot–Savart law and symmetry to show that for all points P above the sheet, and all points P´ below it, the magnetic field B is parallel to the sheet and directed as shown. (b) Use Ampere's law to find B at all points P and P´.

FIRST PART

Vertical ComponentsCancel

Apply Ampere to Circuit

Infinite Extent

B

B

Li

: thereforeis loop theinsideCurrent

lengthunit per current

L

The “Math”

Infinite Extent

B

B

20

0

0

B

LBLBL

id enclosedsB

Bds=0

A Physical Solenoid

Inside the Solenoid

For an “INFINITE” (long) solenoid the previous problem and SUPERPOSITION suggests that the field OUTSIDE this

solenoid is ZERO!

More on Long Solenoid

Field is ZERO!

Field is ZERO

Field looks UNIFORM

The real thing…..

Weak Field

Stronger

Fairly Uniform field

Finite Length

Another Way

0

:

0

0

0

niB

nihBhh

id

Ampere

enclosed

sB

Application

Creation of Uniform Magnetic Field Region

Minimal field outsideexcept at the ends!

Two Coils

“Real” Helmholtz Coils

Used for experiments.

Can be aligned to cancelout the Earth’s magneticfield for critical measurements.

The Toroid

Slightly lessdense than

inner portion

The Toroid

r

NiB

so

totalNirBd

Ampere

2

turns)# (N 2

:nintegratio ofpath the

in contained coil INNER about the

only worry need Weagain.

0

0

sB

15.  A wire with current i=3.00 A is shown in Fig.29-46. Two semi-infinite straight sections, both tangent to the same circle, are connected by a circular arc that has a central angle θ and runs along the circumference of the circle. The arc and the two straight sections all lie in the same plane. If B=0 at the circle's center, what is θ?

38.  In Fig. 29-64, five long parallel wires in an xy plane are separated by distance d=8.00 cm , have lengths of 10.0 m, and carry identical currents of 3.00 A out of the page. Each wire experiences a magnetic force due to the other wires. In unit-vector notation, what is the net magnetic force on (a) wire 1, (b) wire 2, (c) wire 3, (d) wire 4, and (e) wire 5?