Textbook: 8.3, 8.4, 8.5 Homework: pg. 407 #1 – 5 pg. 414 #1 - 4, 6, 9 Magnetic Force on a Conductor, Ampere’s Law & Electromagnetic Induction Textbook: 8.3, 8.4, 8.5 Homework: pg. 407 #1 – 5 pg. 414 #1 - 4, 6, 9 pg. 419 #1, 3, 8 Derive magnetic force on a conductor: One charged particle: F = qvBsin(theta) Many charged particles: F = n(qvBsin(theta)) I = nq/del_t (solve for q) v = l/del_t Substitute in and simplify

FM on a Conductor in a Magnetic field: The magnetic force FM [N] on a conductor of length l [m] carrying a current I [A] through a magnetic field B [T] is: Where  is the angle between I and B RHR: Thumb points in direction of +ve I, fingers in direction of B, palm pushes in direction of FM Pg. 405 #1 - 4

Review - around a conductor: Magnetic field strength is proportional to current: Magnetic field strength is inversely proportional to radius: Therefore:

Ampère’s Law Along any closed path through a magnetic field, the sum of the products of the scalar component of B, parallel to the path segment with the length of the segment, is directly proportional to the net electric current passing through the area enclosed by the path.

Ampere’s Law Take a closed path in B Add up B||l around path Sum equals 0I Permeability of free space 0 = 4 x 10-7 Tm/A Ex. Derive magnetic field of a straight line conductor - Take path as circle (B is constant and always points parallel to path) - Sum(B||*del_l) = Sum(B*del_l) - B*Sum(del_l) = B*2*pi*r = mu* I - B = mu*I/(2*pi*r) Pg. 409 #1 Ex. Derive magnetic field of a solenoid - Take path as rectangle and note: outside solenoid B = 0, and B parallel to solenoid so parts of path equal zero - Only path that remains is inside solenoid in direction of magnetic field - Sum (B||*del_l) = Sum (B*del_l) = B*Sum(del_l) = B*L - Current in rectangle is NI (current from N turns) - B = mu*NI/L Pg. 411 #7

FM between TWO CONDUCTORS Pg 412

Two parallel straight conductors 5 Two parallel straight conductors 5.0 m long and 12 cm apart are to have equal currents. The force each conductor experiences from the other is not to exceed 2.0 x 10-2 N. What is the maximum possible current in each conductor? 49 A

Applications Coaxial Cable (see pg. 410) Definition of Ampere/Coulomb Electric Shielding Magnetic Shielding Definition of Ampere/Coulomb A current of 1 A in parallel wires 1 m apart in a vacuum creates a force of 2 x 10-7 N per meter 1 C is the charge transported by 1 A in 1 s Coaxial Cable Electric Shielding: An electric field diagram will show that the electric field created by the inner cable will terminate on the outer sleeve and will not continue outside. This shields the electric field. (Demo???) Magnetic Shield: Since a current I flows one way and an equal current I flows the other way the total current through a closed path outside the conductor is zero. Therefore the magnetic field is zero. Coaxial cables do not induce magnetic fields outside them. Pg. 413 #9, 10

Maglev Trains Opposite poles levitate train Guide magnets change polarity to push and pull train Very little friction leads to very high speeds (World Record is 581 km/h)

Electromagnetic Induction A changing electric field (i.e. a current) induces a magnetic field A changing magnetic field induces a current in nearby conductors Examples: Electric Generators Transformers Guitar Pickups

Lenz’s Law The direction of an induced current is such that it opposes the changing magnetic field that created it Pg. 418 #2, 3

Magnitude of the magnetic field strength in the core of the solenoid B is the magnitude of the magnetic field strength in the core of the solenoid, in teslas; I is the current flowing through the coil, in amperes; L is the length of the solenoid, in metres; N is the number of turns on the coil.

Pg 418 # 3 Two magnets are dropped through thin metal rings (Figure 11). One of the rings has a small gap. (a) Will both magnets experience a retarding force? Explain your reasoning (b) Will your answers change if the rings are replaced with long cylinders, one with a long thin gap down one side?