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1.) (150 points) A circular wire loop of radius TO is centered at the origin (rctr = (0,0,0)) in the xy-plane and carries a constant current I in the counterclockwise direction looking down on the loop from above (the positive z-axis). (a) (50 points) Using the integral formula I derived in class for the vector potential in the "radiation gauge", i.e., with V. A = 0, and adapted to this filamentary current (I) situation, calculate A(r) for large r = r " ro and show that it is given by the dipole term, namely, A = dl 21 > (6) Here, T = + y y + z Z III r + ZZ and TO = ro (x cos 0 + is 0), is the position vector of a point on the circle at azimuthal angle 0 = tan- The vector m is the magnetic dipole moment of the current loop. In your calculation of A, you are to show that m is given by m m=10 = (7) where a is the area vector of the circular loop, with magnitude a = mr and direction relative to the direction of the current I given by the right-hand-rule; in this case, a = nozz. (Hint: Draw a careful picture showing the current loop and all relevant vectors. Use cylindrical coordinates (T1, 0, 2) (see Eqs. (1)). If you are careful, you will be left with a simple integral to do. Another Hint: Check your units as you calculate! (b) (20 points) Present a plausible argument (not necessarily a mathematical proof), valid for r To, that the result for A in Eq. (6) is valid for any simple (nonintersecting) planar loop whose dimensions are all comparable to TO. (c) (20 points) Calculate the magnetic field B for a dipole m with vector potential A given by Eq. (6). Your answer should be in terms of m and r. This is a good place to use Eqs. (2,3). (d) (30 points) Suppose that the current I in part (a) is due to a particle of charge q and mass M going around the circle of radius To with velocity U = ud. Show that (magnitude and direction!) m = qL 2Mc (8) where L = To X Mv is the (orbital) angular momentum of the particle.

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