Physics Friday 65

Consider a loop of wire, forming a simple closed plane curve. We let the plane of the loop be the xy plane. Then let us parametrize the curve via (x(u),y(u)), 0≤uumaxwith counterclockwise orientation (so that the normal vector to the loop via the right-hand rule is in the positive z direction).

Suppose we then have a uniform magnetic field of magnitude B directed at an angle α from the positive z axis; we can choose our x axis so that is in the xz plane with positive x component.

Now, let us have a current I in the loop (positive I indicates counterclockwise current). What then, is the force on an element du at
(x(u),y(u)), and what is the net effect of this force on the loop?

The magnetic force due to field on a length of wire carrying a current I is . Thus, for an element du, the length vector is . We also have . Thus, we have force
.
To find the total force, we integrate over u:

(here we have used the fact that , as our curve is closed, and similarly for ).

So there is no net force on the loop. However, let us pick a point (xc,yc,0) in the plane of the loop, and find the torque about that point. For an element of the curve, the torque element is
.

Integrating this, and using again the fact that , we get
.

Now,
,
and similarly, , so we see that the x and z components of the torque are zero, and
. Note that the torque is independent of the plane point (xc,yc,0) chosen, so we can choose it to be the center of mass of the loop.

For the remaining integral in our torque, we note that
, and to this integral over our closed curve (c), we apply Green’s Theorem (see also here), which says
, where D is the plane region bound by the simple closed curve ∂D. Here, f(x,y)=0, g(x,y)=x, so
, and so
,
where A is the area enclosed by our loop. Our torque is thus:
.
Now, using the area vector , we see that , so
.
Noting that the magnetic moment of an object can be defined by the torque it experiences from an external magnetic field via
, we see that the magnetic moment of any planar current loop is
. Lastly, notice that the direction of our torque is such that the loop rotates to align the magnetic moment (and thus the area vector) with the magnetic field.

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2 Responses to “Physics Friday 65”

  1. Physics Friday 66 « Twisted One 151’s Weblog Says:

    […] Friday 66 By twistedone151 Last time, we showed that a plane loop carrying current I placed in a uniform magnetic field experiences a […]

  2. Physics Friday 124 « Twisted One 151's Weblog Says:

    […] by the loop. Thus, we obtain , where is the area vector. This is the same result that we found here by considering the torque a constant magnetic field exerts on a current loop. Similarly, if we […]

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