1. Copyright © 2009 Pearson Education, Inc. Ampère’s Law

2. Copyright © 2009 Pearson Education, Inc. The integral 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 I encl flowing through the loop:

3. Copyright © 2009 Pearson Education, Inc. so B = (μ 0 I)/(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

4. Copyright © 2009 Pearson Education, Inc. 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. (c) If R = 2.0 mm & I = 60 A, Calculate B at r = 1.0 mm, r = 2.0 mm, & r = 3.0 mm. Example: Field Inside & Outside a Current Carrying Wire

5. Copyright © 2009 Pearson Education, Inc. Field Due to a Long Straight Wire From Ampere’s Law Outside of the wire, r > R so B = (μ 0 I)/(2πr), as before.

6. Copyright © 2009 Pearson Education, Inc. Inside the wire, r < R Here, we need I encl, the current inside the Amperian loop. The current is uniform, so I encl = (  r 2 /  R 2 )I B(2  r) = μ 0 (  r 2 /  R 2 )I or B = (μ 0 Ir)/(2  R 2 )

7. Copyright © 2009 Pearson Education, Inc. Field Due to a Long Straight Wire: Summary The field is proportional to r inside the wire. B = (μ 0 Ir)/(2  R 2 ) The field varies as 1/r outside the wire. B = (μ 0 I)/(2πr)

8. Copyright © 2009 Pearson Education, Inc. A Coaxial Cable is a single 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. Conceptual Example: Coaxial Cable.

9. Copyright © 2009 Pearson Education, Inc. Solving Problems Using Ampère’s Law Ampère’s Law is most useful for solving problems when there is considerable Symmetry. Identify the Symmetry. Choose an Integration 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. Copyright © 2009 Pearson Education, Inc. Magnetic Field of a Solenoid & 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. B = 0 outside the solenoid, & the path integral is zero along the vertical lines, so the field is (n = number of loops per unit length):

11. Copyright © 2009 Pearson Education, Inc. Example: Field Inside a Solenoid 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.