Week 7: Sources of the Magnetic Field 247
once again, as was neglecting fringe fields for capacitors (the moral equivalent in the electrostatic
case).
That was certainlyveryeasy compared to any sort of Biot-Savart Law integration. The latter
canbe done with some work, but it isn’t easy and requires more calculus than you are likely to have
so far; maybe some day in a future class you’ll do it.
Simple, easy or not, the solenoid is an enormously useful and important example, so be sure you
learn it completely.
Example 7.6.4: Toroidal Solenoid
N turns
a
b
r
C
h
z
I
Figure 89: A cross-sectional view of a toroidal solenoid withN turns, and a rectangular cross-
sectional with inner radiusa, outer radiusb, and heighth, carrying currentIin each turn. The field
both inside the solenoid is concentric to the vertical axis of the torus (from symmetry and the right
hand rule), leading to the Amperian Path shown.
In figure 89 above a toroidal^70 solenoid is drawn. The particular one we will look at has a
rectangular cross-section although (as we will see) this doesn’t really matter as far as finding the
field in all of space is concerned – any uniform cross-sectional shape (such as a circle or ellipse or
outline of Homer Simpson) would do. We choose a rectangle with nice coordinates mostly to make
it easy to compute the self-inductance of this solenoidnextweek, not because it mattersthisweek
and this way we can just reuse the figure as well as the Ampere’s Lawresult.
The wires in the figure (drawn on the left) have to be visualized wrapping the whole torus (fairly
tightly). If one lays one’s right hand thumb mentally along the direction of the current in each leg
of a loop around the torus, you can easily convince yourself thateachwire produces a field nearby
that is generally cylindrically “around” the torus in the direction givenby laying your thumb in
the direction of theinside wires, the ones closes to thez-axis of symmetry. In this case theB-
field is counterclockwise, then, viewed from our perspective above, and our Amperian Path (along
which the field should be constant in magnitude and tangent to the path or anything you like and
perpendicular) is a circle of radiusr.
We locate the circleinsidethe solenoid at first. Ampere’s Law then gives:
∮CB~·d~ℓ = μ 0 Ithru C=μ 0∫
S/CJ~·nˆdAB 2 πr = μ 0 N I
Bt =μ 0 N I
2 πr(542)
where we discover that the current “throughC” is just the current in a single wire times the number
of wires butonly when the curveClies inside the torus! For circlesCoutside of the torus the
(^70) Wikipedia: http://www.wikipedia.org/wiki/Torus. A torus is a “doughnut shape”, usually with a circular cross
section.