ics ri Apr.13. Announcements Magnets arth/dipole ield Charges & Magnetic ields riday, April 13, 2007 Help sessions 9-10 pm in C 119 Masteringics U #20 due Mon., April 16 U #21 due ri., April 20 Rework #2 due ri., April 20 http://www.voltnet.com/ladder/ orksheet Problem #1 Remember when we talked about the motion of charges in a wire. v d t A The charges move with an average velocity v d. Remember when we talked about the motion of charges in a wire. v d t A The magnetic force on a wire with charge carriers moving with velocity v d in a uniform magnetic field should just be the vector sum of the force on each individual charge. ince the average velocity is the same for all charge carriers in a straight wire, the magnetic force acts in the same direction (on average) on all the charge carriers. Therefore... (straight wires) = qvd sin where is the angle between the long dimension of the wire and the magnetic field. The force acts in the direction perpendicular to both. 1
(straight wires) = qvd sin = nal = umber of charge carriers n = number of charge carries per unit volume A = cross-sectional area of wire L = length of wire = nalqvd sin ut recall our definition of current in a wire... ubstituting, we get the simpler epression: = nqv A d = Lsin o, the maimum value of the magnetic force on a current carrying wire occurs when the wire is perpendicular to the magnetic field and has the value... Hey, Mr. luggo. Magnetic orce on a wire is named after me! ma = L ingers point in the direction of the magnetic field. Thumb points in the direction of the current. Palm faces the direction of the magnetic force. orksheet Problem #2 o what happens if we put a loop of wire carrying current in a = Lsin The currents in the top and bottom of this loop are anti-parallel and parallel to the magnetic field. Therefore, sin = 0. o the magnetic forces on the top and bottom of the loop are 0! 2
o what happens if we put a loop of wire carrying current in a = Lsin o what happens if we put a loop of wire carrying current in a = Lsin On the left side of the loop, we use the right-hand rule to determine that the force is out (toward us). On the right side of the loop, we use the right-hand rule to determine that the force is in (away from us). n each case, = 90 o, sin = 1, so = L n each case, = 90 o, sin = 1, so = L o, the magnitude of the force on the left side of the loop is the same as the magnitude of the force on the right side of the loop. net = 0 hat, therefore, is going to happen to the loop in this net = 0 orksheet Problem #3 The loop rotates! The magnetic force produces a TORQU on the current loop, causing it to rotate. n this case, the loop rotates counterclockwise as viewed from above... a/2 a/2 left = d = Ld = b a 2 counter-clockwise b 3
The loop rotates! The magnetic force produces a TORQU on the current loop, causing it to rotate. n this case, the loop rotates counterclockwise as viewed from above... a/2 a/2 right = d = Ld = b a 2 counter-clockwise b The loop rotates! The magnetic force produces a TORQU on the current loop, causing it to rotate. n this case, the loop rotates counterclockwise as viewed from above... a/2 a/2 total = left + right = ab = (Area of loop) b = A sin where is the angle between the normal to the loop and the magnetic field. normal Top View: The normal is the direction perpendicular to the plane of the loop of wire. f the loop has turns, then the torque becomes... orksheet Problem #4 = A sin A small circular coil of 20 turns of wire lies in a uniform magnetic of 0.5 T so that the normal to the plane of the coil makes an angle of with the direction of. The radius of the coil is 4 cm, and it carries a current of 3 A. hat is the magnitude of the torque on the coil?.......... ut what about the forces on the top and bottom parts of this loop? ut what about the forces on the top and bottom parts of this loop? The force on the bottom of the loop is down, trying to epand the loop. The force on the top of the loop is up, also trying to epand the loop. 4
.................... The sum of these two oppositely directed forces, as you might have guessed, is ZRO! otice that the shape of the loop (a circle) only impacts the final result in that the magnitude of the force is proportional to the area the loop encloses. e ve already developed a RHR for the direction of the magnetic force on a current-carrying wire. f we d like to use the vector math to determine the direction, we can use the following for the force on a straight wire. L = L hat if the wire is bent or curved relative to the magnetic field lines (or vice versa)? ds e use our usual trick, dividing the wire into a bunch of little pieces that are straight d = ds hat if the wire is bent or curved relative to the magnetic field lines (or vice versa)? then add up the contributions from all the little pieces along the wire ds = d s. n 1819, Hans Oersted discovered that current carrying wires deflected compass needles! The current in the wire must be generating the magnetic field the compasses are detecting! 5
r ell on what could the magnetic field around a wire depend? certainly it depends upon the current in the wire. the distance from the wire. And some constants... The magnetic field around a very long, straight current carrying wire is: ith the permeability of free space given by μ 0 = μ 0 2 7 = 4 10 Tm / A As increases, should increase. As r increases, should decrease. and where is the perpendicular distance to the wire is the current in the wire. n our earlier eample of Oersted s eperiment... rap your hand around the wire with your thumb pointed in the direction of the current. Your fingers curl around the wire in the direction the magnetic field points.. The compass needles point in the direction of the magnetic field. Let s try the right hand rule on this eample and see if it agrees with the compasses! orksheet Problem #5 orksheet Problem #6 6