Lecture 23: WED 11 MAR Ampere’s law Physics 2102 Jonathan Dowling André Marie Ampère (1775 – 1836)

Exam 02 and Midterm Grade Q1&P1, Dr. Schafer, Office hours: MW 1:30-2:30 pm, 222B Nicholson Q2&P2, Dr. Gaarde, Office hours: TTh 2:30-3:30 pm, 215B Nicholson Q3&P3, Dr. Lee, Office hours: WF 2:30-3:30 pm, 451 Nicholson Q4&P4, Dr. Buth, Office hours: MF 2:30-3:30 pm, 222A Nicholson Exam 02 Average: 63/100 You can calculate your midterm letter grade, as posted on paws, via this handy formula used by all sections: [(MT01+MT02)*500/200+HWT/761]*30]/530*100 = Midterm Score Here MT01 and MT02 are your scores on the first and second midterms and HWT is your total homework points for HW01-07 only. Once you have this Midterm Score you then can get your letter grade via: A: B: C: D: F: below 50. Details on the grading policy are at: I have posted both midterm exam grades and your midterm class score on WebAssign.

Magnetic field due to wire 1 where the wire 2 is, a I2I2 I1I1 L Force on wire 2 due to this field, F Forces Between Wires Neil’s Rule: Same Currents – Wires Attract! Opposite Currents – Wires Repel! Neil’s Rule: Same Currents – Wires Attract! Opposite Currents – Wires Repel!

Summary Magnetic fields exert forces on moving charges: The force is perpendicular to the field and the velocity. A current loop is a magnetic dipole moment. Uniform magnetic fields exert torques on dipole moments. Electric currents produce magnetic fields: To compute magnetic fields produced by currents, use Biot- Savart’s law for each element of current, and then integrate. Straight currents produce circular magnetic field lines, with amplitude B=  0 i/2  r (use right hand rule for direction). Circular currents produce a magnetic field at the center (given by another right hand rule) equal to B=  0 i  /4  r Wires currying currents produce forces on each other: Neil’s Rule: parallel currents attract, anti-parallel currents repel.

Given an arbitrary closed surface, the electric flux through it is proportional to the charge enclosed by the surface. q Flux = 0! q Remember Gauss Law for E-Fields?

New Gauss Law for B- Fields! No isolated magnetic poles! The magnetic flux through any closed “Gaussian surface” will be ZERO. This is one of the four “Maxwell’s equations”. There are no SINKS or SOURCES of B-Fields! What Goes IN Must Come OUT! There are no SINKS or SOURCES of B-Fields! What Goes IN Must Come OUT!

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. 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. i1i1 i2i2 i3i3 ds i4i4 Thumb rule for sign; ignore i 4 If you have a lot of symmetry, knowing the circulation of B allows you to know B. Ampere’s law: Closed Loops

Sample Problem Two square conducting loops carry currents of 5.0 and 3.0 A as shown in Fig What’s the value of ∫B∙ds through each of the paths shown? Path 1: ∫B∙ds =   (–5.0A+3.0A) Path 2: ∫B∙ds =   (–5.0A–5.0A–3.0A)

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. (Go around loop in same direction as B field lines!) R Much Easier Way to Get B-Field Around A Wire: No Cow-Culus.

Ampere’s Law: Example 2 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) 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) i Current out of page, circular field lines

Ampere’s Law: Example 2 (cont) Current out of page, field tangent to the closed amperian loop For r < R For r>R, i enc =i, so B=  0 i/2  R = LONG WIRE! For r>R, i enc =i, so B=  0 i/2  R = LONG WIRE! Need Current Density J!

R r B O Ampere’s Law: Example 2 (cont)

Solenoids The n = N/L is turns per unit length.