Composite Conductor [20 marks]
A long cylindrical conductor (a wire),
viewed in cross-section at right,
carries a uniform current density out of the page
except in the off-centre cylindrical hole
where the metal has been removed.
The radius of the hole is half the radius R of the wire
and the hole's central axis is located its R/2 to the left
of the central axis of the wire, as shown.
If R = 1 cm and the wire carries a net current of 1 A,
calculate the resultant magnetic field at a position P
on the x axis 1 cm to the right of the wire's right edge.
- (a)
- [5 marks] Explain how you plan to solve this problem.
ANSWER: Treat the wire as a superposition of
an intact cylinder (without the hole) carrying a current I0
plus a smaller, off-centre cylinder (the hole) carrying a current I1
in the opposite direction. The resultant field is
the superposition of the fields due to those two components.
- (b)
- [15 marks] Do it.
ANSWER: The area of the hole is 1/4 that of the whole wire, so the actual
current I is only 3/4 of I0, the current that the intact wire
would be carrying at the same current density. Thus
I0 = 4I/3
and
I1 = - I/3. Both components produce a vertical (on this page)
magnetic field at P, the first upward (in the same direction as the
R arrow on the diagram) and the other downward. The two fields
are given simply by Ampère's Law for a long cylindrically
symmetric current distribution (outside the conductor):
giving
.
(The direction has already been specified.)
Comment: As announced at the beginning of the hour,
this was a long and difficult exam. I did not expect anyone to
successfully complete every part to every question; nor did I
expect anyone to be bored! I hope everyone had a chance to at
least look at the selection of questions and work first on those
with which they felt most confident. Now that the exam is over,
I encourage everyone to go back and do the other questions,
the ones that looked scariest when you first saw them.
I can think of no better way to boost your competence (and confidence)
for the final exam, which is coming soon . . . .
Meanwhile, I will be devising an algorithm for scaling the marks
on this Midterm.