no tagsLecture 026: The Application of Ampere's LawSteveSekula, 1 November 2010 (created 31 October 2010)GoalsDemonstrate the application of Ampere's LawDiscuss the magnetic field due to a solenoid, and its applicationsApplication: Solar CurrentsShow the movie of the solar flares again.Solar storms consist of large currents of electrically charged particlesmoving in the sun's magnetic fields. Consider a rectangular Amperian Loopthat has long dimension and is in the presence of a constantmagnetic field whose strength is . What is the total currentenclosed by the loop?According to our picture (Wolfson Figure 26.30b), two legs of the loop liein the direction of the magnetic field, so . For the other two, thedot product is zero. Thus:This is then equal to:General Physics - E&M (PHY 1308) LectureNotesGeneral Physics - E&M (PHY 1308) LectureNotesL 00 0 m = 4 Â 19B 0 T = 2 Â 1À3r I IB~Á d~ = Ö0B r dr ~Á d~ = Br dr B r BL IB~Á d~ = 2ZB = 2Zd = 2General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...1 of 7 11/01/2010 08:34 AMWe can then solve for the current:I 0 AConceptual Puzzle: Direction of Currents in WiredAmpere's Law is also useful for solving for the direction of unknowncurrents in wires. Consider three parallel wires, labeled A, B, and C, inwhich equal magnitude current flows in the same direction in two of thembut in the opposite direction in the third (you don't know which is which atfirst). Consider two loops, one which encircles A and B (Loop 1) and onewhich encircles B and C (Loop 2). If the following is true:Then:What is around Loop 1?ANSWERIf the above is true about Loop 2, it means that the net currentenclosed is NONZERO and current in B and C must flow in the samedirection. That means that because current must beopposite in one of the three wires to the flow of the other two, and thatmeans current enclosed would be ZERO.1.Which wire carries the current opposite the other two?SOLUTIONAs argued above, it must be wire A.2.2BL I = Ö0= 2Ö0BL= 2(4Ù 0 N=A )Â 1À72(2:0 0 T)(400 0 m)Â 1À3Â 19= 112r = (Loop 2) IB~Á d~ = 0r HB~Á d~r (Loop 1) HB~Á d~ = 0General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...2 of 7 11/01/2010 08:34 AMApplication: Magnetic Field in a Long WireConsider an infinite straight wire along the x-axis. We've previously treatedthe wire as thin and calculated the magnetic field outside the wire. Wefound it to be:Bwhere is the distance from the wire.Let's consider a long, straight wire carrying a constant current, . The wirenow has a thickness; let's treat is as a cylinder whose radius is . Werecognize that this problem has line symmetry. This means that we expectthat no matter what, the magnetic field will depend only on the radialdistance from the center of the wire (the cylinder axis).Outside the wire, we expect the magnetic field lines to circle around thewire. We can choose our Amperian loops to take advantage of this fact, andmake them also circles centered on the wire axis. What about inside thewire? If we draw a circular Amperian loop inside the wire, it will enclose afraction of the current but that current will still be a cylindrical flow withcircular magnetic field lines surrounding it. This in all cases in thisproblem:where is the radius of the enclosed cylinder.Let's consider the two big cases in this problem: outside the wire, wherethe enclosed current is fixed and equal to , and inside the loop where theenclosed current is a fraction of the total current (to be determined).Outside the wireHere, the enclosed current is no matter how far we are from the wire.outside=Ö I02Ùyy I R ` d` (2Ùr) IB~Á d~=IB = Br I I General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...3 of 7 11/01/2010 08:34 AMThus:and thus:Bwhere , our distance from the center of the wire. We've simplyrecovered the result from the Biot-Savart Law!Inside the wireInside the wire, our Amperian Loops only enclose a fraction of the totalcurrent. How do we figure it out?If the current, , is constant and uniformly distributed through the wire,then the current density is constant in the wire. Thus:for the whole wire is the same as:for a current enclosed in a loop whose radius is . Thus:Thus:2ÙrB I = Ö0=2ÙrÖ I0r = yI J =A =(ÙR ) = I = I2J =(Ùr ) = Jenclosed= Ienclosed2r J ! = JenclosedÀ Ienclosed= Ir2R2General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...4 of 7 11/01/2010 08:34 AM2ÙrB Iand thus:BIn both cases, application of the right-hand rule tells us that the magneticfields circulate counter-clockwise when viewing the current coming at us.The SolenoidOne of the most important applications of a current-carrying wire is thesolenoid. Take a straight wire and start winding it around a rod or a tubeto make a tight coil. Remove the rod/tube. The remaining current-carryingcoil generates a magnetic field that is nearly uniform inside the coil andweak outside the coil.Why is this useful? Many applications need a uniform, well-characterizedmagnetic field (just like applications like capacitors and particleaccelerators benefit from a uniform electric field). Consider MRI machines,where the patient needs to be immersed in a strong, uniform magneticfield. They are actually being inserted into a large solenoid.Let's analyze the solenoid as an extension of the current loop. Consider themagnetic dipole field resulting from a single current loop. Now add asecond loop next to it, then a third. inside the coil, the field becomes moreuniform. Outside it grows weaker and weaker. For an infinitely longsolenoid, the field inside will be perfectly uniform and the field outside willbe zero.Real solenoids are not perfect,, and do have weak an non-uniform electricfields outside the coil. We can analyze the magnetic field of a solenoid veryclose to the center of a long coil, so that we are far from the ends and thecoil is effectively infinite in either direction away from the center.Draw a cross-section of a coil, carrying a current . Now draw arectangular Amperian Loop whose long dimension is that encloses inside= Ö0r2R2inside=Ö Ir02ÙR2I L N General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...5 of 7 11/01/2010 08:34 AMturns for the coil. For every additional coil enclosed, the Amperian Loopencloses an additional of current.
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