## Transcribed Text

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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