# If torque is a form of moment, then why is the word 'moment' used for the 'magnetic dipole moment', NiA, for a current-carrying loop?

I thought torque is a form of moment: If the moment makes the object rotate, we call it torque; otherwise we just use the name moment.

However it seems the electric & magnetic dipole moments are different. Why are we calling just the quantity $$NiA$$, the moment?

A coil (of area $$A$$ and $$N$$ turns, carrying current $$i$$) in a uniform magnetic field $$\vec B$$ will experience a torque $$\vec\tau$$ given by $$\vec\tau = \vec\mu \times \vec B.$$ Here $$\vec\mu$$ is the magnetic dipole moment of the coil, with magnitude $$\mu=NiA$$ and direction given by the right-hand rule.

The word "moment" is used widely within physics (and also in certain parts of mathematics). The overall thread that unites all the various uses of this word is that you typically have a quantity $$Q$$ which is located some distance $$d$$ away from the center of interest, in which case you can define the first moment of $$Q$$ as $$M=Qd.$$ It is possible to extend this to higher-order moments, such as $$M_n=Qd^n$$, and to quantities which are distributed over an axis as $$Q(x)$$, in which case we need to integrate in something of the form $$M_n=\int_{-\infty}^\infty Q(x) x^n \mathrm dx$$.

Prominent examples of this are electric and magnetic dipole moments (as well as their generalizations to quadrupole, octupole, and other multipole moments), gravitational multipole moments, moments of inertia, and so on.

Torque is a form of moment — specifically, moment of force. Because of its ubiquity in certain ways of interfacing with physics, sometimes it is common to drop the "of force" qualifier, and just use the word "moment" as synonymous with "torque". But it is important not to lose sight that this is shorthand for a longer concept.

And, as a general rule, it is a good idea not to use the word "moment" without specifying what it's a moment of, or otherwise saying what kind of moment it is. Thus, we call the quantity $$NiA$$ the magnetic dipole moment, and while we might sometimes drop the "magnetic" if it is clear from the context, we never refer to it as just "moment" without preceding it with "dipole".

And, while we're here:

If the moment makes the object rotate, we call it torque; otherwise we just use the name moment

No. If you're using the word moment (of force) for torque, the identification holds regardless of circumstances.

• so dipole moment = $NiA$, here Q := Ni and $d^n := A$ ? very clear explanation appreciate it very much thanks! Commented Oct 29, 2022 at 16:37
• also doesn't torque require the object to rotate? as torque causes change in angular momentum.. because i haven't seen the usage of word torque in statics problems like calculating bending moment etc.. Commented Oct 29, 2022 at 16:38
• Using "torque" for a moment that causes twisting or rotation about an axis, and "moment" for a moment that causes bending, is more an engineering convention than a physics one. As far as I know there is no strict rule beyond convention, as the two things I described are physically the same thing (moment of force). Commented Oct 29, 2022 at 16:47
• Regarding the magnetic dipole moment, Wikipedia explains in more depth. As an initial model, if you have a current $I$ circulating on a wire, you can define the magnetic dipole moment of the loop as the vector line integral $\vec\mu = \oint \vec r \times I\mathrm d\vec\ell$, where you're really taking the (vector cross product) first moment of the current density. For a flat loop, you can then calculate the integral explicitly as the area vector of the loop. Commented Oct 29, 2022 at 16:55
• @across Indeed, the magnetic dipole moment can be thought of as a moment of current. But that description is incomplete, because magnetostatics cares about more moments of current than just the first-order dipolar one (so you will often need e.g. the magnetic quadrupole moment of the distribution, which (i) has a quadratic polynomial instead of a linear factor in the integral, and (ii) is not a vector but a tensor, basically because there are many more ways to form quadratic combinations of the components of $\vec r=(x,y,z)$ than there are linear combinations. Commented Oct 29, 2022 at 20:04