Example: Magnetic field from a long wire

Force between two current-carrying wires

Magnetic field from a circular current loop

Magnetic field in terms of dipole moment

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Category: $mathematics$ mathematics

Magnetic Field

1. Magnetic Field

Electron current
Biot-Savart Law
Lines of Magnetic Field
Ampere’s Law
Solenoids and Toroids

2. Electron current

i=n/t

3. Detecting magnetic field

https://www.compadre.org/osp/EJSS/4500/285.htm
Simple simulation of magnetic field

4. Sources of Magnetic Fields

Magnetic fields exert forces on moving charges.
Something reciprocal happens: moving charges give rise to
magnetic fields (which can then exert a force on other
moving charges).
We will look at the easiest case: the magnetic field created
by currents in wires.
The magnetism of permanent magnets also comes from
moving charges (the electrons in the atoms).

5. Magnetic Interaction

Rather than discussing moving charges in general,
restrict attention to currents in wires. Then:
A current generates a magnetic field.
A magnetic field exerts a force on a current.
Two conductors, carrying currents, will exert forces on each other.

6. Biot-Savart Law

The mathematical description of the magnetic field B due
to a current-carrying wire is called the Biot-Savart law. It
gives B at a selected position.
A current I is moving all through the wire. We need to add
up the bits of magnetic field dB arising from each
infinitesimal length dl.
dl
I
q
r
dB

7. Biot-Savart Law

8. Biot-Savart Law

9. Biot-Savart Law

dl
I
q
r
dB
The constant m0 = 4p x 10-7 T m/A
is called the permeability of free space.
IT TURNS OUT THAT M 0 AND E O ARE RELATED IN A SIMPLE
WAY: ( E 0 M 0 ) -1/2 = 3X10 8 M/S = C, THE SPEED OF LIGHT.
WHY? LIGHT IS A WAVE OF ELECTRIC AND MAGNETIC
FIELDS.

10. Example: Magnetic field from a long wire

Consider a long straight wire carrying a current I.
We want to find the magnetic field B at a point P,
a distance R from the wire.
P
R
I

11. Example: Magnetic field from a long wire

Consider a long straight wire carrying a current I.
We want to find the magnetic field B at a point P,
a distance R from the wire.
x
dx
dl
r
x
P
0
R
I
Break the wire into bits dl.
To do that, choose coordinates:
let the wire be along the x axis,
and consider the little bit dx at a
position x.
The vector r = r ^r is from this bit
to the point P.

12. Example: Magnetic field from a long wire

x
dx
q
Direction of dB: into page.
r
x
0
R
I
+

13. Example: Magnetic field from a long wire

x
dx
q
r
x
0
R
I
Direction of dB: into the page.
This is true for every bit; so we
+ don’t need to break into
components, and B also points
into the page.

14. Example: Magnetic field from a long wire

x
dx
q
r
x
0
R
I
Direction of dB: into the page.
This is true for every bit; so we
+ don’t need to break into
components, and B also points
into the page.
Moreover, lines of B go around a
long wire.

15. Example: Magnetic field from a long wire

Moreover, lines of B go around a
long wire. Perspective:
x
dx
q
r
x
i
0
R
I
+
B
P
Another right-hand rule

16. Example: Magnetic field from a long wire

x
dx
q
Direction of dB (or B): into page
r
x
0
R
I
+

17. Example: Magnetic field from a long wire

x
dx
q
r
x
0
R
I
+

18. Force between two current-carrying wires

B2
d
I1
B1
I2
Current 1 produces a magnetic
field B1 =m0I/ (2p d) at the position
of wire 2.
This produces a force on current 2:

19. Force between two current-carrying wires

B2
d
I1
B1
I2
Current 1 produces a magnetic
field B1 =m0I/ (2p d) at the position
of wire 2.
This produces a force on current 2:
F2 = I2L x B1

20. Force between two current-carrying wires

B2
d
I1
F2
B1
I2
Current 1 produces a magnetic
field B1 =m0I/ (2p d) at the position
of wire 2.
This produces a force on current 2:
F2 = I2L x B1

21. Force between two current-carrying wires

B2
d
I1
F2
B1
I2
Current 1 produces a magnetic
field B1 =m0I/ (2p d) at the position
of wire 2.
This produces a force on current 2:
F2 = I2L x B1
This gives the force on a length L of wire 2 to be:
Direction: towards 1, if the currents are in the same direction.

22. Force between two current-carrying wires

Current I1 produces a magnetic
field B1 =m0I/ (2p d) at the position
I1 of the current I2.
B2
d
F2
B1
I2
This produces a force on current I2:
F2 = I2L x B1
Thus, the force on a length L of the conductor 2 is given by:
[Direction: towards I1]
The magnetic force between two parallel wires
carrying currents in the same direction is attractive .
What is the force on wire 1? What happens if one current is reversed?

23. Magnetic field from a circular current loop

along the axis only!
B
Only z component
is nonzero.
dBperp
r
dBz
z
a
dl
R
I

24. Magnetic field from a circular current loop

along the axis only!
B
dBperp
r
z
a
dl
At the center of the loop
At distance z on axis
from the loop, z>>R
dBz
R
I

25. Magnetic field in terms of dipole moment

Far away on the axis,
B
The magnetic dipole moment of the
loop is defined as m = IA =IpR2.
The direction is given by the right
hand rule: with fingers closed in
the direction of the current flow,
the thumb points along m.
z
m
R
I

26.

Magnetic field in terms of dipole moment
In terms of m, the magnetic field on axis (far from
the loop) is therefore
This also works for a loop with N turns. Far from
the loop the same expression is true with the
dipole moment given by m=NIA = IpNR2

27. Dipole Equations

Electric Dipole
t=p x E
U = -p·E
Eax = (2pe0 )-1 p/z3
Ebis = (4pe0 )-1 p/x3
Magnetic Dipole
t=m x B
U = -m·B
Bax = ( m0/2p) m/z3
Bbis = (m0/4p) m/x3

28. Ampere’s Law

Electric fields
Coulomb’s law gives E
directly (as some integral).
Magnetic fields
Biot-Savart law gives B
directly (as some integral).
Gauss’s law is always true. It Ampere’s law is always true.
is seldom useful. But when it It is seldom useful. But when
is, it is an easy way to get E. it is, it is an easy way to get B.
Gauss’s law is a surface
integral over some
Gaussian surface.
Ampere’s law is a line integral
around some
Amperian loop.

29. Ampere’s Law

Draw an “Amperian loop”
around the sources of
current.
The line integral of the
tangential component of B
around this loop is equal to
moIenc:
I2
I3
Ampere’s law is to the Biot-Savart law exactly
as Gauss’s law is to Coulomb’s law.

30. Ampere’s Law

Draw an “Amperian loop”
around the sources of
current.
The line integral of the
tangential component of B
around this loop is equal to
moIenc:
The sign of Ienc comes
from another RH rule.
I2
I3
Ampere’s law is to the Biot-Savart law exactly
as Gauss’s law is to Coulomb’s law.