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Magnetic Field Sources
Recall that:
(1) Magnetic fields are caused by currents.
(2) Currents in magnetic fields can experience magnetic forces.
(Also recall that a “current” can be a single charged particle in
motion with some velocity v.)
In the previous chapter we focused on item (2), the forces
experienced by currents (or by individual moving charges) in a
magnetic field. In this chapter we focus on (1), the calculation of
several magnetic field configurations caused by currents (or by
point charges in motion).
The magnetic field of a single point charge
Here is the picture of the magnetic field created as a charge q moves
with velocity v. As you can see, it’s a bit complicated.
The B field circulates
around the particle’s
direction of travel.
Try out the right hand
rule!
B decreases in
strength with
increasing distance
from the charge.
v, r, and B appear to
involve a cross
product relationship.
P is the point
where the B
field is being
calculated.
The magnetic field of a single point charge
This B field is described by the Biot-Savart Law.
 k m qv  rˆ
B
r2
with
km 
0
4
This is the magnetic equivalent of Coulomb’s Law.
Notice the similarities and differences:
Jean-Baptiste Biot
1774-1862
It has a force constant, km , in place of the Coulomb
constant, k, and, like the electric force, it is proportional
to q. Also, like the electric force, it decreases as the
square of the distance!
As “expected”, the B vector depends on the cross
product of v with r.
“Permeability”
of free space.
 0  4  10 7
Tm
A
k m  10 7
Tm
A
Félix Savart
1791-1841
Magnetic fields superpose. Example:
Find the direction and
magnitude of the magnetic
field at the origin in terms of
the given quantities.
Biot-Savart Law for an element of current
This looks like the field of a point charge in motion, but with qv replaced
by Idl. The Biot-Savart Law for current can be derived from the point
charge equation, by first putting that equation into infinitesimal form:
 km (dq)v  rˆ
dB 
r2
Work on the infinitesimals:



dl dq 
(dq )v  dq

dl  Idl
dt dt
 The Biot-Savart Law for an
element of current:

 km Idl  rˆ
dB 
r2
Magnetic field of a line of current.
(We looked at this last chapter.)
This B field has cylindrical
symmetry. Compare and
contrast to E field of line
charge.
Magnetic field of a line of current, from the
Biot-Savart Law for an element of current
Notice that at point P the magnetic field from all
current elements along the line is pointed in the
same direction—into the page. The sine factor
below comes only from the cross product.

 km Idl  rˆ
dB 
r2
x
sin(  ) 
r
and
dB 
km I sin(  )
dl
2
r
dl  dy
dB 
km Ix
dy
3
r
Integrating along line from –a to +a:
a
a
km Ix
dy
B   3 dy  km Ix 
2
2
r
x

y
a
a


3/ 2
B
For infinitely long wire, a >> x. Here is the B field with
cylindrical symmetry, falling off as “1/r” from the wire:
2km aI
x x2  a2
2k m I
B
x
Magnetic fields superpose. Another example:
2k m I
B
x
Can be used at each location Pi to find
the total field at that point by
superposing the fields from both wires.
Magnetic dipole field of a current loop
Magnetic field on the axis of a current loop (single turn)
Notice that at point P the magnetic field
element components dB due to dl “add” in
the x direction and cancel in the transverse
directions as we integrate around the loop.

 km Idl  rˆ
dB 
r2
cos( ) 
a
r
km I cos( )
dB 
dl
r2
dB 
km Ia
dl
3
r
Integrating around the loop from 0 to 2:
2a
B

0
k m Ia
k m Ia
dl

r3
r3
2a
k m Ia
2k mIa 2
2k mIa 2
2k m 
dl

(
2

a
)



3/ 2
3/ 2
0
r3
r3
x2  a2
x2  a2
At the center of the loop, x = 0:

B

2k mI  0 I

a
2a


Magnetic field on the axis of a current loop
(for N turns we multiply B by N)
Total force between two point charges (as they pass)
This case is exceptionally simple since, at the
time these two charges pass each other, the
cross products are evaluated at 90o. So we will
be finding magnitudes.
The particle at the origin creates electric and
magnetic fields felt by the other particle:
E
kq
r2
 k m qv  rˆ
B
r2
B
km qv
r2
The particle at y = r is moving to the left at speed –v.
It experiences a total force due to both fields:


 
F  qE  qv  B
F  qE  qvB
Putting the E and B equations into F gives:
2
2
2
2
 km qv  kq  km v  kq  v 
 kq 
  2 1  2 
F  q 2   qv 2   2 1 
k  r  c 
r 
 r  r 
Looks like a
magnified
electric field!
Discuss.
Magnetic force between parallel currents
How does the force depend on whether the currents are
going in the same direction or in opposite directions???
Magnetic force between parallel currents
Magnetic field due to current I:
0 I
B
2r
Force felt by I’ in this magnetic field:
0 I  0 II L

F  I LB  I L

2r
 2r 
If the two currents are equal:
0 I 2 L
F
2r
0 I 2 L
F
2r
How could you use such a device to measure current?
Ampere’s Law
André-Marie Ampère – 1775-1836
On September 11, 1820 Ampere heard of H.
C. Ørsted's discovery that a magnetic needle
is acted on by a voltaic current. Only a week
later, on September 18, he presented a
paper containing a far more complete
exposition of that and other phenomena.
The SI unit of measurement of electric
current, the ampere, is named after him.
Ampère's fame mainly rests on the service
that he rendered to science in establishing
the relations between electricity and
magnetism, and in developing the science of
electromagnetism, or, as he called it,
electrodynamics.
Ampere’s Law
Recall, for a moment, the electric field calculations we did earlier. Coulomb’s
Law was the basis for these calculations, but for problems involving symmetric
systems it was easier to use Gauss’s Law whenever possible.
The situation is similar for magnetic field calculations. The Biot-Savart Law
describes sources of magnetic fields at the most fundamental level. But for
systems with symmetric magnetic fields, Ampere’s Law, which we shall now
introduce, can greatly simplify the calculation. We’ll write down the equation first,
then interpret it, and do a number of examples:
 
 
 B  ds  0  J  dA  0 I encl
The integral of B around a closed loop
is proportional to the total current
enclosed by the loop.
Recall that Gaussian surfaces are chosen for
purposes of calculation. So are “Amperian loops”
B
A
J
Magnetic field of a line of current.
This time, we use Ampere’s Law.
Ampere’s Law:
 
 B  ds  0 I encl
Integrate around any circle
centered on wire, in direction of
B. Cylindrical symmetry means
we can bring B outside integral:
B(2r )   0 I
Solve for the magnetic field:
 0 I 2k m I
B

2r
r
Simple!
Magnetic field of a solenoid
Field approximately
uniform inside, and
close to zero outside.
S
N
Magnetic field of a solenoid, from Ampere’s Law
Ampere’s Law:
 
 B  ds  0 I encl
Using the integration path
shown, the only contribution to
the integral comes from the
segment a b:
BL  0 NI
Solve for the magnetic field:
B
0 NI
L
 0 nI
N = number of turns in loop
Simple!
n = turns per unit length in
the solenoid
Magnetic field on the axis of a solenoid: realistic field
In a real solenoid:
(1) The central field is
slightly lower than the
value given by Ampere’s
Law.
(2) The end field is exactly
half the Amperean value.
Note: This curve cannot
be found by using
Ampere’s Law. It must
be found by integrating
over a “stack of loops”.
Magnetic field of a toroid
A donut is an example of a “torus”
(so is a coffee cup). A “toroid” is a
solenoid that has been bent into a
circle, with its ends connected to
form a torus.
The magnetic field is contained
inside the toroid, and the field lines
are circles centered at the center
of the torus. But, unlike a
solenoid, the field is not uniform: it
varies with r.
Ampere’s Law:
 
 B  ds  0 I encl
The magnetic field will be constant around
any circle inside the toroid. The total current
passing through any such circle is NI. So:
B(2r )  0 NI
Solve for the magnetic field:
0 NI 2k m NI
B

2r
r
Magnetic field for cylinder with uniform current density
We already know the magnetic field outside
the cylinder. The enclosed current is I, so
the field is the same as that for a wire (by
cylindrical symmetry).
For the field inside the cylinder, we
first must find the current density:
I
I
J 
A R 2
So at r, the enclosed current is:
I encl  JAencl
Ir 2
 I  2
  2 (r )  2
R
 R 
And Ampere’s Law gives:
B(2r )  0 I encl 
0 Ir 2
R2
 0 Ir
B
2R 2
Magnetic fields of sheets of current
1. Sketch magnetic fields for
finite sheets of current.
2. Extend to infinite sheets of
current.
3. Find magnetic field near
an infinite sheet of current.
Find: the magnetic field for a coaxial conductor
Find: the magnetic moment of a hydrogen atom
Magnetic field
Magnetic field