Ampère’s Law Figure 28-8. 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. Ampère’s Law relates the magnetic field B around a closed loop to the total current Iencl flowing through (& perpendicular to) the loop: Figure 28-8. 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.

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 28-9. Circular path of radius r. so B = (μ0I)/(2πr), as before.

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.

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.

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

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)

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)

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.

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.

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):

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.