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≤*u*≤*u*_{max}with 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 (*x*_{c},*y*_{c},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 (*x*_{c},*y*_{c},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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Tags: Electric Current, Electricity & Magnetism, Friday Physics, Green's Theorem, Magnetic Moment, physics, Torque

This entry was posted on March 27, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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April 3, 2009 at 1:21 am |

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

June 18, 2010 at 12:23 am |

[…] 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 […]