Lesson 21
(1) Circulation
Definition of Circulation: Consider a closed curve in a region where a magnetic field
is present. Assign a direction of going round the curve. Divide the curve into small
segments each so small that they can be approximated by small displacement
vectors d in the assigned direction. The
circulation of the magnetic field is defined as
ò B × d = åB× d
number of segments ® ¥
Example: In a vortex, the velocity runs counter
clockwise with magnitude proportional to the
distance r from the center:
u = wr w = constant
The circulation of the velocity field around a circle of radius r in the counterclockwise direction is
ò u×d
= u ´ 2p r = 2pwr 2
Example: Calculate the circulation of a uniform magnetic field in the positive xdirection around the semicircular loop on the x-y plane in a clockwise direction as
shown.
Solution: For the current element at an
angle q from the x-axis, the length is
d = adq
and the direction is clockwise.
B × d = Bd cos (p 2 - q ) = Basinq dq
The line integral along the semicircular arc is
p
Ba ò sinq dq = 2Ba
0
Along the diameter,
B × d = Bdx
The line integral is
B ò dx = -2aB
-a
a
ò B×d
Therefore
=0
(2) Ampere’s Law
For a closed curve on a plane, once a direction of going
around is chosen, the direction of the normal associated
with it is then fixed to be the direction of advance of a right
hand screw when it is turned in the direction of going
around the curve.
Ampere’s law states that for any closed curve,
ò B×d
= m0 å Iin
where I in is any current flowing through the
interior of the closed curve, and is considered
positive if it is in the direction of the normal
associated with sense of going around the
closed curve. Otherwise it is negative.
We can verify this law using the magnetic field of an infinitely long current carrying
wire. If the closed curve is chosen to be a circle of radius r around the wire, and the
direction of going around is in the same direction of the magnetic field on the curve,
then
ò B×d
=
2p
m0 I
ò 2p r rdq = m I
0
0
If the closed curve is chosen to consist of two circular arcs of radii r1 and r2 joined
together by radial straight segments as shown, then with the sense of going around
chosen to be counter-clockwise when the current is
out of the paper, the line integrals are:
ò B×d
=
m0 I
mI
r2 Dq = 0 Dq
2p r2
2p
=
ò B×d
ab
ò B×d
bc
=0
because B is
da
perpendicular to the straight segments
ò B×d
cd
=-
m0 I
mI
r1Dq = - 0 Dq
2p r1
2p
Therefore
ò B×d
=0
Alternatively, we can use Ampere’s law to derive the magnetic field due to the
infinitely long wire. In this approach, we first note that because of symmetry, the
magnetic field lines of the wire are expected to be circles around the wire, and the
field strength is the same along each circle. Choosing such a circle of radius r as the
closed curve to apply Ampere’s law, we see that the circulation is
ò B×d
= B ( 2p r )
Ampere’s law then gives 2p rB = m0 I
relative to the direction of the current.
B=
m0 I
2p r
with the correct direction
(3) Magnetic field of a thick wire
An infinitely long wire with circular cross-section of radius a has the same
cylindrical symmetry as a thin wire if the current density j inside it depends only
on the distance r from the central axis. Choosing again a circular loop of radius r to
calculate the circulation of magnetic field, two cases now occur.
Case 1: r ³ a . The current inside the loop is the total
current I : Iin = I . Just as for a thin wire, Ampere’s law
leads to the magnetic field
B=
m0 I
2p r
Case 2: r < a . To calculate I in , the cross-section of the
wire is divided into thin rings of thickness dr¢with
radius r¢ . The area of the ring is 2p r¢dr¢. The current in
the ring is dI = 2p jr¢dr¢ . Therefore
I in =
r
ò 2p jr¢dr¢
0
In the special case when the current density is
uniform, j is a constant, and is given by
I
j= 2
pa
I
Thus, I in = 2p
p a2
ærö
ò r¢ dr¢ = çè a ÷ø I
0
r
2
ærö
Ampere’s law gives
2p rB = ç ÷ I
èaø
A plot of B against r is as shown.
2
so that
B=
m0 Ir
2p a 2
(4) Magnetic field of a long solenoid
By symmetry, the magnetic field of an infinitely long solenoid is parallel to the axis
of the solenoid. The field lines are therefore straight lines parallel to the axis.
Symmetry also requires the field strength to be constant along each field line.
We choose for the application of
Ampere’s law a rectangular loop with
two of its sides parallel to the axis. There
is no contribution to the circulation of
magnetic field along this loop from the
two sides perpendicular to the axis.
Along the sides parallel to the axis, in the
case when the loop is outside the
solenoid, the contribution is
B1 - B2
where is the length of the sides and B1, B2 the magnetic fields on the two sides.
Since no current exists inside the loop, Ampere’s law leads to the conclusion
B1 = B2
which states that the field outside the solenoid is uniform. Expecting the field to be
zero far away from the solenoid, we conclude that the field is zero outside the
solenoid.
For the field inside the solenoid, we choose the loop to straddle the solenoid as
shown. In this case, the contribution to the circulation is entirely from the side
inside the solenoid:
ò B×d
=B
The current inside is given by
Iin = n I
where n is the turn density. Ampere’s law then leads to
B = m0 nI
Note that the direction of the magnetic field relative to the current flow in the
solenoid can be obtained from the right hand: fingers aligned with current and
thumb with magnetic field inside.
(5) Magnetic field of toroid
By symmetry, field lines are expected to be circles centered at the axis of symmetry
of the toroid and field strength is constant along each such lines. Choosing for the
application of Ampere’s law such a field line, we see two cases. In the first case,
where the field line is outside the toroid, the total current inside the loop is zero. We
therefore conclude that the magnetic
field is entirely confined to the
interior of the toroid. In the second
case, the field line loop is chosen to be
inside the toroid. The current inside
the loop is now
Iin = NI
where N is the number of turns.
Ampere’s law then gives
B=
m0 NI
2p R
where R is the distance from the axis of symmetry. Thus, unlike the solenoid, the
field strength is not uniform inside a toroid.
(6) Magnetic field of current sheet
Current flows in the x-direction of an infinite sheet coinciding with the x-y plane.
The magnetic field at a point above the sheet ( z > 0) is in the positive y-direction
while that at a point below ( z < 0) is in the negative y –direction. This can be shown
by dividing the sheet current into pairs symmetrically placed about the x-axis. The
magnetic field of such a pair has no component perpendicular to the sheet by
symmetry.
Symmetry also implies that the field
strength above is equal to that below as
long as the two positions are
equidistant from the sheet. Choosing for
the application of Ampere’s law a
rectangular loop straddling the sheet as
shown, with the sides parallel to the
sheet at equal distance from the sheet,
taken to be of length , we find
ò B×d
= 2B
The current inside is
Iin = l
where l is the density of surface current,
and has the unit of A / m . From Ampere’s
law,
1
B = m0 l
2
Note that the magnetic field is discontinuous across a current sheet.