Physics 2102
Jonathan Dowling
Physics 2102
Lecture 16
Ampere’s law
André Marie Ampère
(1775 – 1836)
Ampere’s law:
Remember Gauss’ law?
Given an arbitrary closed surface, the electric flux through it is
proportional to the charge enclosed by the surface.
q
Flux=0!
r r q
$ # ! E " dA =
%0
Surface
q
Gauss’ law for Magnetism:
No isolated magnetic poles! The magnetic flux through any closed
“Gaussian surface” will be ZERO. This is one of the four
“Maxwell’s equations”.
B
•
d
A
=
0
!
r r q
E
"
d
A
=
!
#0
Ampere’s law:
a Second Gauss’ law.
"
r r
B ! d s = µ0 i
i4
loop
The circulation of B
(the integral of B scalar
ds) along an imaginary
closed loop is proportional
to the net amount of current
traversing the loop.
i2
i3
i1
ds
r r
# B " d s = µ 0 (i1 + i2 ! i3 )
loop
Thumb rule for sign; ignore i4
As was the case for Gauss’ law, if you have a lot of symmetry,
knowing the circulation of B allows you to know B.
Sample Problem
• Two square conducting loops carry currents
of 5.0 and 3.0 A as shown in Fig. 30-60.
What’s the value of ∫B·ds through each of
the paths shown?
Ampere’s Law: Example 1
• Infinitely long straight wire
with current i.
• Symmetry: magnetic field
consists of circular loops
centered around wire.
• So: choose a circular loop C
-- B is tangential to the loop
everywhere!
• Angle between B and ds = 0.
C
(Go around loop in same
direction as B field lines!)
R
r r
! B " ds = µ0i
C
Bds
=
B
(
2
"
R
)
=
µ
i
0
!
µ 0i
B=
2!R
Ampere’s Law: Example 2
i
• Infinitely long cylindrical
wire of finite radius R
carries a total current i with
uniform current density
• Compute the magnetic field
at a distance r from
cylinder axis for:
– r < a (inside the wire)
– r > a (outside the wire)
r
Current into
page, circular
field lines
R
r r
B
"
d
s
=
µ
i
0
!
C
Ampere’s Law: Example 2 (cont)
r r
B
"
d
s
=
µ
i
0
!
C
B(2!r ) = µ 0ienclosed
µ 0ienclosed
B=
2!r
ienclosed
r
Current into
page, field
tangent to the
closed
amperian loop
2
i
r
2
= J (!r ) = 2 !r = i 2
!R
R
µ 0ir
B=
2
2!R
2
For r < R
For r>R, ienc=i, so
B=µ0i/2πR
Solenoids
r r
" B • ds = µ0ienc
r r
" B • ds = 0 + B h + 0 + 0
ienc = iN h = i ( N / L)h = inh
r r
" B • ds = µ0ienc ! Bh = µ0inh ! B = µ0in
Magnetic Field of a Magnetic Dipole
A circular loop or a coil currying electrical current is a magnetic
dipole, with magnetic dipole moment of magnitude µ=NiA.
Since the coil curries a current, it produces a magnetic field, that can
be calculated using Biot-Savart’s law:
r
r
r
µ0
µ0 µ
µ
B( z ) =
"
2
2 3/ 2
2! ( R + z )
2! z 3
All loops in the figure have radius r or 2r. Which of these
arrangements produce the largest magnetic field at the
point indicated?
Sample Problem
Calculate the magnitude and direction of the
resultant force acting on the loop.