Today’s agenda: Ampere’s Law.

1. Today’s agenda: • Ampere’s Law. • You must be able to use Ampere’s Law to calculate the magnetic field for high-symmetry current configurations. • Solenoids. • You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s Law to make numerical magnetic field calculations for solenoids and toroids.

2. ´ dl I Magnetism The Laws of Biot-Savart & Ampere

3. Fundamental Law for Calculating Magnetic Field • Biot-Savart Law (“brute force”) • Ampere’s Law (“high symmetry”) • Example: Calculate Magnetic Field of ¥ Straight Wire • from Biot-Savart Law • from Ampere’s Law • Calculate Force on Two Parallel Current-Carrying Conductors

4. "Brute force" "High symmetry" Calculation of Electric Field • Two ways to calculate the Electric Field: • Coulomb's Law: • Gauss' Law • What are the analogous equations for the Magnetic Field?

5. ´ "Brute force" I "High symmetry" Calculation of Magnetic Field • Two ways to calculate the Magnetic Field: • Biot-Savart Law: • Ampere's Law • These are the analogous equations for the Magnetic Field!

6. dl X r q dB I Biot-Savart Law…bits and pieces (~1819) So, the magnetic field “circulates” around the wire

7. P r R q x I dx Þ Þ \ Þ Magnetic Field of ¥ Straight Wire • Calculate field at point P using Biot-Savart Law: • Rewrite in terms of R,q: Which way is B?

8. P r R q x I dx Þ \ Magnetic Field of ¥ Straight Wire

9. dl I R • Evaluate line integral in Ampere’s Law: • Apply Ampere’s Law: Þ Magnetic Field of ¥ Straight Wire • Calculate field at distance R from wire using Ampere's Law: ´ • Choose loop to be circle of radius R centered on the wire in a plane ^ to wire. • Why? • Magnitude of B is constant (fcn of R only) • Direction of B is parallel to the path. • Current enclosed by path =I • Ampere's Law simplifies the calculation thanks to symmetry of the current! ( axial/cylindrical )

10. F ´ Ib d L Ia Þ Force on b = Ib d L Ia ´ Þ Force on a = F Force on 2 ParallelCurrent-Carrying Conductors • Calculate force on length L of wire b due to field of wire a: • The field at b due to a is given by: • Calculate force on length L of wire a due to field of wire b: The field at a due to b is given by:

11. Ampere’s Law Just for kicks, let’s evaluate the line integral along the direction of B over a closed circular path around a current-carrying wire. I ds B r The above calculation is only for the special case of a long straight wire, but you can show that the result is valid in general.

12. The current I passing through a loop is positive if the direction of integration is the same as the direction of B from the right hand rule. Ampere’s Law I is the total current that passes through a surface bounded by the closed (and not necessarily circular) path of integration. Ampere’s Law is useful for calculating the magnetic field due to current configurations that have high symmetry. I I positive I negative I ds B B ds r r

13. Your text writes because the current that you use is the current “enclosed” by the closed path over which you integrate. Your starting equation sheet has The reason for the 2nd term on the right will become apparent later. Ignore it for now. If your path includes more than one source of current, add all the currents (with correct sign). I1 ds I2

14. Example: a cylindrical wire of radius R carries a current I that is uniformly distributed over the wire’s cross section. Calculate the magnetic field inside and outside the wire. I R Cross-section of the wire:  direction of I B R r

15. Over the closed circular path r:  direction of I B R r Solve for B: B is linear in r.

16. B R r Outside the wire:  direction of I R A lot easier than using the Biot-Savart Law! r (as expected). B Plot:

17. Today’s agenda: • Ampere’s Law. • You must be able to use Ampere’s Law to calculate the magnetic field for high-symmetry current configurations. • Solenoids. • You must be able to use Ampere’s Law to calculate the magnetic field of solenoids and toroids. You must be able to use the magnetic field equations derived with Ampere’s Law to make numerical magnetic field calculations for solenoids and toroids.

18. Magnetic Field of a Solenoid A solenoid is made of many loops of wire, packed closely together. Here’s the magnetic field from a loop of wire: Some images in this section are from hyperphysics.

19. Stack many loops to make a solenoid: Ought to remind you of the magnetic field of a bar magnet.

20.             B              I    l You can use Ampere’s law to calculate the magnetic field of a solenoid. N is the number of loops enclosed by our surface.

21.             B              I    l Magnetic field of a solenoid of length l, N loops, current I. n=N/l (number of turns per unit length). The magnetic field inside a long solenoid does not depend on the position inside the solenoid (if end effects are neglected).

22. A toroid* is just a solenoid “hooked up” to itself. Magnetic field inside a toroid of N loops, current I. The magnetic field inside a toroid is not subject to end effects, but is not constant inside (because it depends on r). *Your text calls this a “toroidal solenoid.”

23. Example: a thin 10-cm long solenoid has a total of 400 turns of wire and carries a current of 2 A. Calculate the magnetic field inside near the center.

24. “Help! Too many similar starting equations!” long straight wire use Ampere’s law (or note the lack of N) center of N loops of radius a probably not a starting equation solenoid, length l, N turns field inside a solenoid is constant solenoid, n turns per unit length field inside a solenoid is constant toroid, N loops field inside a toroid depends on position (r) “How am I going to know which is which on the next exam?”

25. Ampere’s Circuital Law in Integral and Differential Form