Physics 9
Spring 2012
NAME:
TA:
Quiz 5 - Solutions
Make sure your name is on your quiz, and please box your final answer. Because we will
be giving partial credit, be sure to attempt all the problems, even if you don’t finish them!
The current density in a particle beam with circular cross section of radius a points along
the beam axis with a magnitude that decreases linearly with J0 at the center r = 0 to half
that value at the edge r = a.
1. Find a linear function J (r) that describes the particle beam.
2. Find an expression for the total current in the beam.
————————————————————————————————————
Solution
1. We are told the the current density decreases linearly with radius, and so the density
must be a function
J (r) = J0 − cr,
where c is a constant that we can determine using the given boundary conditions. The
chosen function already sets J (0) = 0, while J (a) = 21 J0 , and so
1
J0
J0 = J0 − ca ⇒ c = .
2
2a
Hence, our current density function is
r
.
J (r) = J0 1 −
2a
2. The current is the flux of the current density,
Z
I = J dA.
For a (radially symmetric) circular beam, the area element dA = 2πrdr, and so
Z a
I = 2π
rJ (r) dr.
0
Integrating the result we found in part 1 gives
Z a
2
Z a
r
r2
a
a2
2π
I = 2πJ0
1−
r dr = 2πJ0
r dr −
= 2πJ0
−
=
J0 a2 .
2a
2a
2
6
3
0
0
Thus, we find that the current in this particle beam is
2π
I=
J0 a2 ,
3
which is a factor of 2/3 smaller than the current of a uniform current density beam
(which would have a current of I = J0 × A = πa2 J0 ).
1