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MIT: 8.02 Electricity and Magnetism Help Course PSet 7 (Due Apr 4) Problems Ampere's Law in Action A long coaxial cable consists of two concentric conductors. The inner conductor is a cylinder with radius , and it carries a current uniformly distributed over its cross section. The outer conductor is a cylindrical shell with inner radius and outer radius . It carries a current that is also uniformly distributed over its cross section, and that is opposite in direction to the current of the inner conductor. Calculate the magnetic eld as a function of the distance from the axis. Use the coordinate system described in the gure below. Symbolic Check For the symbolic check, write your answer using some or all of the following: , , R_1 for , R_2 for , R_3 for and mu_0 for . For : (mu_0*I)/(2*pi*(R_1)^2)*r Answer: mu_0*I*r/(2*pi*R_1^2) Direction: φ^ Answer: For : (mu_0*I)/(2*pi*r) Answer: mu_0*I/(2*pi*r) Direction: φ^ For Answer: : mu_0*I*(R_3^2-r^2)/((2*pi*r)*(R_3^2-R_2^2)) Answer: mu_0*I/(2*pi*r)*(1-(r^2-R_2^2)/(R_3^2-R_2^2)) Direction: φ^ For : Answer: 0 Answer: 0 Direction: None, B is zero Answer: None, B is zero Solution: Consider the diagram below, showing a cross-section through the cable. Assume the current ows “out of the page" in the core and “into the page" in the shell. The cylindrical symmetry of the system tells us that the magnetic eld lines will be circles centered on the cable axis, and the magnetic eld strength will depend only on distance from the axis. So we may construct an Amperian loop of radius centered on the cable axis as shown, knowing that will always be tangent to the loop (in the constant magnitude around the loop. direction) and will have a Applying Ampere's law to our loop with counterclockwise circulation, we nd where is the component of zero component of in the counterclockwise azimuthal direction (the only non- ). Our task now is to nd for the various regions of interest. (a) The current in the core is distributed uniformly throughout its cross-sectional area of . An Amperian loop of radius encloses an area of , and will therefore enclose a fraction of this total current given by the ratio of these two areas: . Plugging this into our expression for , we have (Note that we have regarded the enclosed current as positive, since it ows in the same direction as the normal to the surface bounded by our Amperian loop, given our choice of counterclockwise circulation.) (b) An Amperian loop of radius in the region will enclose all of the current in the core and none of the current in the shell, so we simply have (c) The outer shell carries a current (negative because it is directed opposite to the normal to the surface bounded by our counterclockwise Amperian loop) which is distributed uniformly throughout its cross-sectional area of . A loop in the region will enclose of shell cross-section, and thus will enclose a fraction of the shell current given by the full to get . The loop also encloses of the core, so in this region . Plugging in : (d) Finally, in the region , our loop encloses the full the shell. The enclosed current is thus , giving of the core and the full of (Note: when we say “loop encloses current ", strictly speaking we mean “current passes through an open surface bounded by loop ".) Just to get a better feel for the dependence of on , we can take , and plot , as shown below. is plotted in units of and in units of . Note that is continuous everywhere. Discontinuities in only come from sheets of current, just as discontinuities in only come from sheets of charge. Answers are displayed within the problem © All Rights Reserved