## Posts Tagged ‘Magnetic Moment’

### Physics Friday 124

June 18, 2010

Part 4: Distant Fields

Let us consider the first-order magnetic potential and field far from a localized current density . The first step is to consider the denominator of the integral .
In particular, choosing an origin within the localized distribution, so that , we expand the fraction to first order in r’, obtaining
,
giving us the expansion
.

Now, we need to use a bit of vector calculus. For any well-behaved scalar field u(r) and vector field v(r), then vector integration by parts over a volume V bounded by simple surface S
.
Now, letting our vector field be the current density J, if our volume V completely encloses (and extends beyond) the region to which the current is localized, then J, and thus uJ, is thus zero on the surface S, and so that integral remains zero as the volume is expanded to all space, leaving us
,
or combining,
.
Now, if we let , and use the magnetostatic requirement that , then we see that
,
,
where Jx is the x component of the current density vector field. Note that this integral is the x component of the integral first integral in our expansion, . Similarly, using  and  each give us

and
,
respectively. Thus, the initial term in our expansion, the monopole term, is zero. Now, indexing the cartesian components of our vectors with i=1,2,3, we now have
.
Now, we return to our integration by parts formula. We use , and 
(where  is the unit vector for the ith coordinate); together with our requirement that the current be divergence-free, we obtain

Now, taking our integral term , and expanding the dot product into the sum of coordinate products via index j, we have
.
Now, since , we can subtract half of this integral from each integral in the above sum, and
,
giving us


Now, for i=j, the above integral is zero. Similarly, we note that ; and similar expressions, which allow us to write
,
where  is the Levi-Civita symbol. This means
;
but this is the ith component of a cross product (see here), so
we have
.
Now, we define the magnetic moment density by , and it’s integral . Then, in terms of this, we see
.
Taking the curl of this, and using the product rule for curl,
, along with the fact that μ is a constant vector, we see
,
where r=|r| is the distance from the origin, and  is the unit vector in the direction of r. Note that the field of an electric dipole of dipole moment p is  (see here). Thus, our field is that of a magnetic dipole, and μ is our magnetic moment (thus justifying our terming  as magnetic moment density).
Note that if our current is confined to a current I in a closed plane curve C with line element , then our integral for magnetic moment becomes
,
and via Green’s theorem (reversing our use of it here), we see
, where n is the normal to the plane of the loop (with direction via the right-hand rule), and the A over which we integrate is the region enclosed 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 have point charges with masses mi and charges qi, in motion with positions ri and velocities vi, then (approximating it as a static current distribution), we have
, and so the magnetic moment integral becomes a sum via the delta functions:
. However, the angular momentum of the ith particle is
,
so we obtain
.
If the particles all have the same charge-to-mass ratio , then it factors out of the sum, and
,
where L is the total angular momentum. Note that for electrons, this then gives , the same result we obtained here for a single electron in a circular orbit.

### Physics Friday 66

April 3, 2009

Last time, we showed that a plane loop carrying current I placed in a uniform magnetic field experiences a torque , where  is the magnetic moment of the loop, and  is the area vector for the loop, as given using the right-hand rule.

Now, let us consider an electron moving in a circular orbit of radius r and angular velocity ω0. Then the period of the orbit is , and so the charge passing through any point on the orbit per unit time is , and so we can treat the circular orbit as a current loop of current  (the minus sign indicates that the current is opposite in direction to the motion of the electron, as the electron has a negative charge). Thus, the orbit has a magnetic moment, .

Unlike the previous case of a current loop, here we also have to consider the angular momentum of our electron’s orbit as well. As our orbit is circular, the angular momentum is just , where me is the mass of the electron. Thus, we can rewrite the magnetic moment in terms of the angular momentum:
.
and the magnetic moment is proportional to the angular momentum of the orbit, with the constant of proportionality dependent only on the properties of the electron.

Note here that since the magnetic moment is proportional to the angular momentum, the torque due to an external magnetic field,  is perpendicular to both  and . Thus, as seen in previous work, the orbit, if not perpendicular to the magnetic field, will precess about the field. This is an example of Larmor precession, which occurs whenever there is a magnetic moment proportional to angular momentum exposed to an external magnetic field. The frequency of this precession, the Larmor frequency, for this problem is . In the more general form, the constant of proportionality between the magnetic moment and angular momentum is called the gyromagnetic ratio (or sometimes the magnetogyric ratio), and usually denoted by γ: . Then the Larmor frequency is . These frequencies, when applied to a charged particle with spin, are important in spin transitions, and play an important role in systems such as nuclear magnetic resonance

### Physics Friday 65

March 27, 2009

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.