1. Ampère’s Law
Figure Arbitrary path enclosing a current, for Ampère’s law. The path is broken down into segments of equal length Δl.

2. The sum is taken around the outside edge of the closed loop.
Ampère’s Law relates the
magnetic field B around a
closed loop to the total
current Iencl flowing through
(& perpendicular to) the loop:
Figure Arbitrary path enclosing a current, for Ampère’s law. The path is broken down into segments of equal length Δl.
The sum is taken around the
outside edge of the closed loop.

3. so B = (μ0I)/(2πr), as before.
Example: Use Ampère’s
Law to find the field around a
long straight wire. Use a
circular path with the wire at
the center; then B is tangent
to dl at every point. The
integral then gives: =
Figure Circular path of radius r.
so B = (μ0I)/(2πr), as before.

4. Example: Field Inside & Outside a
Current Carrying Wire
A long, straight cylindrical wire
conductor of radius R carries a
current I of uniform current density
in the conductor. Calculate the
magnetic field due to this current at
(a) points outside the conductor (r > R) (b) points inside the conductor (r < R).
Assume that r, the radial distance
from the axis, is much less than
the length of the wire.
Solution: We choose a circular path around the wire; if the wire is very long the field will be tangent to the path.
a. The enclosed current is the total current; this is the same as a thin wire. B = μ0I/2πr.
b. Now only a fraction of the current is enclosed within the path; if the current density is uniform the fraction of the current enclosed is the fraction of area enclosed: Iencl = Ir2/R2. Substituting and integrating gives B = μ0Ir/2πR2.
c. 1 mm is inside the wire and 3 mm is outside; 2 mm is at the surface (so the two results should be the same). Substitution gives B = 3.0 x 10-3 T at 1.0 mm, 6.0 x 10-3 T at 2.0 mm, and 4.0 x 10-3 T at 3.0 mm.

5. Example: Field Inside & Outside a
Current Carrying Wire
Calculate the magnetic
field due to this current at:
(a) points outside the
conductor (r > R)
(b) points inside the conductor
(r < R).
(c) If R = 2.0 mm & I = 60 A,
Calculate B at r = 1.0 mm,
r = 2.0 mm, & r = 3.0 mm.
Solution: We choose a circular path around the wire; if the wire is very long the field will be tangent to the path.
a. The enclosed current is the total current; this is the same as a thin wire. B = μ0I/2πr.
b. Now only a fraction of the current is enclosed within the path; if the current density is uniform the fraction of the current enclosed is the fraction of area enclosed: Iencl = Ir2/R2. Substituting and integrating gives B = μ0Ir/2πR2.
c. 1 mm is inside the wire and 3 mm is outside; 2 mm is at the surface (so the two results should be the same). Substitution gives B = 3.0 x 10-3 T at 1.0 mm, 6.0 x 10-3 T at 2.0 mm, and 4.0 x 10-3 T at 3.0 mm.

6. Field Due to a Long Straight Wire From Ampere’s Law
Outside of the wire, r > R: =
So B = (μ0I)/(2πr), as before.

7. Iencl = (r2/R2)I so B(2r) = μ0(r2/R2)I which gives:
Inside of the wire, r < R: =
Here, we need Iencl, the current inside the
Amperian loop. The current is uniform, so
Iencl = (r2/R2)I so B(2r) = μ0(r2/R2)I which gives:
B = (μ0Ir)/(2R2)

8. Field Due to a Long Straight Wire: Summary
The field is proportional
to r inside the wire.
B = (μ0Ir)/(2R2)
The field varies as 1/r
outside the wire.
B = (μ0I)/(2πr)

9. Solving Problems Using Ampère’s Law Calculate the Enclosed Current.
Ampère’s Law is most useful for solving
problems when there is considerable Symmetry.
Identify the Symmetry.
Choose a Path that reflects the Symmetry
(typically, the best path is along lines where the
field is constant & perpendicular to the field
where it is changing).
Use the Symmetry to determine the direction of
the field.
Calculate the Enclosed Current.

10. Conceptual Example: Coaxial Cable.
A Coaxial Cable is a wire surrounded by a cylindrical
metallic braid. The 2 conductors are separated by an
insulator. The central wire carries current to the other end
of the cable, & the outer braid carries the return current &
is usually considered ground.
Calculate the magnetic field (a) in the space between the
conductors & (b) outside the cable.
Solution: a. Between the conductors, the field is solely due to the inner conductor, and is that of a long straight wire.
b. Outside the cable the field is zero.

11. Magnetic Field of a Solenoid & of a Toroid
Solenoid  A coil of wire with many loops. To find the field
inside, use Ampère’s Law along the closed path in the figure.
Figure 28-16: Cross-sectional view into a solenoid. The magnetic field inside is straight except at the ends. Red dashed lines indicate the path chosen for use in Ampère’s law.
B = 0 outside the solenoid, & the path sum is zero
along the vertical lines, so the field is (n = number of
loops per unit length):

12. Example: Field Inside a Toroid
A thin ℓ = 10 cm long solenoid has a total of N = 400 turns (n = N/ℓ) of wire & carries a current I = 2.0 A.
Calculate the magnetic field inside near the center.
A Toroid is similar to a solenoid, but it is bent into the shape of a circle as shown.
Example: Toroid
Use Ampère’s Law to calculate the magnetic field
(a) inside &
(b) outside a toroid.
Solution: Substitution gives B = 1.0 x 10-2 T.