Chapt. 31
Faraday’s Law of Induction and Inductors
239
evaluate it in terms of
B
,
ω
,
θ
, the cylindrical radial coordinate
r
, and the differential
dz
.
Note that there are two cases. The differential bit of contour on the side of the contour
counterclockwise from ˆ
n
has velocity
−
rω
ˆ
n
and that on the side of the contour clockwise
from ˆ
n
has velocity in the direction
rω
ˆ
n
.
HINT:
Note that ˆ
z
·
dvectors
=
dz
where
dz
could be
implicitly positive for negative.
c) To absorb the two cases for the two sides of ˆ
n
in a single formula let’s introduce coordinate
y
which is in the plane of the contour and perpendicular to the
z
axis. We let
y>
0 for the
counterclockwise side and
y <
0 for the clockwise side. We can now right
±
r
as
y
. Give
the formula for the differential motional emf now.
d) We want to integrate the differential emf over the contour by summing up the contributions
to each
dz
interval first about each height
z
first and then summing these up for the total.
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 Fall '06
 Buchler
 Physics, Magnetic Field, Summation, Clockwise, Law of Induction and Inductors

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