PHYS 1444 – Section 003 Lecture #17AnnouncementsExample 28 – 2Operational Definition of Ampere and CoulombAmpére’s LawSlide 6Verification of Ampére’s LawSlide 8Example 28 – 4Slide 10Example 28 – 5Solenoid and Its Magnetic FieldSolenoid Magnetic FieldSlide 14Example 28 – 8Biot-Savart LawExample 28 – 9Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu1PHYS 1444 – Section 003Lecture #17Monday, Oct. 31, 2005Dr. Jaehoon Yu•Example for Magnetic force between two parallel wires•Ampére’s Law•Solenoid and Toroid Magnetic Field•Biot-Savart LawToday’s homework is homework #9, due noon, next Thursday!!Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu2Announcements•Reading assignments–CH28 – 7, 28 – 8, and 28 – 10 •The 2nd term exam–Date: Monday, Nov. 7–Time: 1 – 2:20pm–Location: SH 103–Coverage: CH 26 – whichever chapter we get to by Wednesday, Nov. 2•Your textbooks–UTA bookstore agreed to exchange your books with the ones that has complete chapters•You need to provide a proof of purchase–Receipts, copy of cancelled checks, credit card statement, etc.Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu3Example 28 – 2 Suspending a current with a current. A horizontal wire carries a current I1=80A dc. A second parallel wire 20cm below it must carry how much current I2 so that it doesn’t fall due to the gravity? The lower has a mass of 0.12g per meter of length. Which direction is the gravitational force? This force must be balanced by the magnetic force exerted on the wire by the first wire.Downward gFl=2I =Solving for I2( ) ( )( )( )( )2 372 9.8 0.12 10 0.20154 10 80m s kg mAT m A App--�״=� � �mgl=MFl=01 22I Idmp0 12mgdl Ipm=Which direction should the current flow? The same direction as I1.Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu4Operational Definition of Ampere and Coulomb•The permeability of free space is defined to be exactly •The unit of current, ampere, is defined using the definition of the force between two wires each carrying 1A of current and separated by 1m –So 1A is defined as: the current flowing each of two long parallel conductors 1m apart, which results in a force of exactly 2x10-7N/m.•Coulomb is then defined as exactly 1C=1A.s.•We do it this way since current is measured more accurately and controlled more easily than charge.704 1 0 T m Am p-= � �Fl=01 22I Idmp=74 101 12 1T m AA Ampp-� ��=72 10 N m-�Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu5Ampére’s Law•What is the relationship between magnetic field strength and the current?–Does this work in all cases?•Nope! •OK, then when?•Only valid for a long straight wire•Then what would be the more generalized relationship between the current and the magnetic field for any shape of the wire?–French scientist André Marie Ampére proposed such a relationship soon after Oersted’s discovery02IBrmp=Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu6Ampére’s Law–The sum of all the products of the length of each segment and the component of B parallel to that segment is equal to 0 times the net current Iencl that passes through the surface enclosed by the path– –In the limit l 0, this relation becomes– 0 enclB l ImD =�P0 enclB dl Im� =�rr�Ampére’s Law•Let’s consider an arbitrary closed path around the current as shown in the figure.–Let’s split this path with small segments each of l long.Looks very similar to a law in the electricity. Which law is it?Gauss’ LawMonday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu7Verification of Ampére’s Law–We just verified that Ampere’s law works in a simple case–Experiments verified that it works for other cases too –The importance, however, is that it provides means to relate magnetic field to current 0 enclIm =•Let’s find the magnitude of B at a distance r away from a long straight wire w/ current I–This is a verification of Ampere’s Law–We can apply Ampere’s law to a circular path of radius r.B =Solving for BB dl� =�rr�Bdl =��B dl =��2 rBp02enclIrmp=02IrmpMonday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu8Verification of Ampére’s Law–How do you obtain B in the figure at any point?•Vector sum of the field by the two currents–The result of the closed path integral in Ampere’s law for green dashed path is still 0I1. Why?–While B in each point along the path varies, the integral over the closed path still comes out the same whether there is the second wire or not.•Since Ampere’s law is valid in general, B in Ampere’s law is not just due to the current Iencl.•B is the field at each point in space along the chosen path due to all sources–Including the current I enclosed by the path but also due to any other sourcesMonday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu9Example 28 – 4 Field inside and outside a wire. A long straight cylindrical wire conductor of radius R carries a current I of uniform current density in the conductor. Determine the magnetic field at (a) points outside the conductor (r>R) and (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=2.0mm and I=60A, what is B at r=1.0mm, r=2.0mm and r=3.0mm? Since the wire is long, straight and symmetric, the field should be the same at any point the same distance from the center of the wire.Since B must be tangent to circles around the wire, let’s choose a circular path of closed-path integral outside the wire (r>R). What is Iencl?Solving for BSo using Ampere’s lawenclI I=0Im =B dl� =�rr�2 rBp02IBrmp=Monday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu10Example 28 – 4 Solving for BSo using Ampere’s lawenclI =20rIRm� �=� �� �B dl� =�rr�2 rBpB =For r<R, the current inside the closed path is less than I. How much is it?What does this mean?The field is 0 at r=0 and increases linearly as a function of the distance from the center of the wire up to r=R then decreases as 1/r beyond the radius of the conductor.22rIRpp=2rIR� �� �� �202I rr Rmp� �=� �� �022IrRmpMonday, Oct. 31, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu11Example 28 – 5 Coaxial cable. A coaxial cable is a single wire surrounded by a cylindrical metallic braid, as shown in the figure. The two
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