W9_parallel_resonance.eps

212 Week 6: Moving Charges and Magnetic Force

The dimensions of this particular loop areaandb, although in the next section we’ll see that
these particular dimensions, and indeed the shape of the plane loop,are not terribly important. I
put the loop in the “inset” to the upper left so you can visualize what itmight look like lying on a
table. Note that we’ll imagine that the loop has an “axle” on which it canfreely pivot. This too
isn’t strictly necessary (we can pick other piviots that will work justas well or better) but guessing
that your recollection of torque is still a bit shaky it won’t hurt to draw in a simple one that is easy
to understand.

In the main part of the figure I’ve drawn an “edge view” of the loop asit sits in auniform
magnetic fieldB~pointing to the right. The “uniform” bit is very important – we obviously would
get a very different result for the force if (for example) the field onthe upperbside were larger than
the field on the lowerbside!

Evaluating the force on each of the four sides of the rectangle is trivial. The upper and lowerb
sides are perpendicular to theB~field, have lengthb, haveNturns each carryingI, and hence the
magnitude of the force is:
Fb=N IbB (450)

We can find the direction easily using the right hand rule. It is up on upper side (with current
pointing out of the page) and down on the lower side.

The force on theasides is hardly more difficult. Let’s consider the one closest to us in thefigure,
with the current slanting down and to the right. The directed current makes an angle ofφwith the
magnetic field, so the force on it is:

Fa=|N I~a×B~|=N IaBsin(φ) (451)

with a direction (right hand rule again) of out of the page. The hiddenaside on the other side (where
the current slants up and to the left) has the same magnitude force and the opposite direction.

The sum of these forces is this clearly

F~tot= (Fb−Fb)yˆ+ (Fa−Fa)zˆ= 0 (452)

where I’ve fairly arbitrarily popped a coordinate system onto the picture withxto the right,yup,
andzout of the page.

Does this (F~= 0) mean that nothing interesting happens to the loop in the field? Not at all!
The twoFaforces are indeed uninteresting, as they act along the same line (along the axle, in fact)
and exert neither force nor torgue on the system. The twoFbforces, however, donotact along the
same line. They exert atorqueon the loop!

How large a torque? Recalling that~τ=~r×F~where~ris a vector from the pivot to the force,
the torque from the upperbside using the pivot shown (so thatr=a/2) is:

τb=a

2

Fbsin(θ) (453)

and pointsinto the page. The torque from the lowerbside is identical in magnitude andhas the
same direction(into the page). The total torque thus has magnitude:

τ=aFbsin(θ) =N I(ab)Bsin(θ) (454)

into the page.

Now take a moment to look carefully at the geometry of this figure. The angleθwe used is the
one between the direction ofa/2 in each case andFb. I’ve drawn this angle in for the lower side
to make it easy to see, but it is the same for the upper side too. If you followθfrom the angle in
between to the angle in the right triangle with the dashed side, useφ=π/ 2 −θ, you can see that
the angle between theright handed normal to the plane loopnˆdrawn and the magnetic field will