This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 44. (a) The ideal solenoid is long enough (and we are evaluating the ﬁeld at a point far enough inside)
such that the open ends of the solenoid are “out of sight” and the situation displays a horizontaltranslational symmetry (assuming the axis of the cylindrical shape of the solenoid is horizontal). A
view of a “slice” of, say, the bottom of the solenoid would therefore appear similar to that shown
in Fig. 30-52, where point P is in the interior of the solenoid and point P is outside the coil. Now,
Fig. 30-52 diﬀers in at least one respect from our “slice” view of the solenoid in that the ﬁeld at
P would be zero instead of what is shown in that ﬁgure. The ﬁeld vanishes there because the
top of the solenoid (similar to that shown in Fig. 30-52, in “slice” view, but with the currents and
ﬁeld directions reversed) would contribute an equal and opposite ﬁeld to any exterior point, thus
canceling it. For interior points, the top and bottom “slices” each contribute 1 µ0 λ (in the same
direction) [this is shown in the solution to problem 39] and thus produce an interior ﬁeld equal to
B = µ0 λ .
(b) Applying Ampere’s law to a rectangular path which passes through points P (interior) and P
(exterior) similar to that described in the solution to part (b) of problem 39, we are not surprised
B · ds = BP − BP ·ˆ x = µ0 λ∆x
just as we found in part (b) of problem 39 (except that we are now taking the +x direction in the
same direction as the ﬁeld at P , to avoid confusion with signs). The diﬀerence with the previous
solution is that in 39, BP − BP ·ˆ was equal to B − (−B ) = 2B , whereas in this case we have
B − 0 = B . Although the value of B is diﬀerent in the two problems, we see that the change
BP − BP ·ˆ is the same: µ0 λ. ...
View Full Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08