I have a problem with determining the direction of torque due to magnetic moment on a current loop. Since the torque is defined as

$$ \tau = p_m \times B $$

so I guess the direction of the torque should be perpendicular to both $p_m$ and $B$.

And yet, take this problem. If a current is flowing through a rectangle loop, as in the following picture, the torque should be directed parallel to the horizontal, and not have any effect in oscillating the loop around it's position of stable equilibrium. (Right?) The axis of rotation is the horizontal line passing through the center, parallel to the sides Look at Fig. 6.14.

But if we sum up the Ampere forces, we find that this torque indeed induces the oscillation of the loop.

I am very confused with the magnetic moment, as it is explained for now. Whenever I come across a problem involving it, there is no picture describing that vector, just the calculation part. I need to know not only it's intensity, but also direction, to fully understand it.


With a constant current, the magnetic (dipole) moment is simply $\bf{m} = \it{I}\bf{a}$ where $\bf{a}$ is the vector area $\bf{a} = \oint\bf{r} \times d\bf{l}$ where $d\bf{l}$ is the differential element around the boundary of the area.

Where the torque $\bf{\tau} = \bf{m} \times \bf{B} $

So for example if the loop is flat like yours above, the direction will be normal to area as per the standard right hand rule. If it is some other shape then the direction will be given by the integral.

This is discussed nicely in Griffiths' E/M

  • $\begingroup$ I agree with you that the magnetic moment will be perpendicular to the flat loop and it's direction determined by the right hand rule... but what about the direction of the torque vector? It should be pointed horizontally then (since B is vertical and m normal to the plane). But then it could not be the cause of rotation... $\endgroup$ – Knight1805 Jan 13 '16 at 20:18
  • $\begingroup$ Remember the direction of the torque vector is always normal to the direction of rotation so if the torque vector is pointed out of the page (like it is in the picture), the loop rotates around it counterclockwise, grab the vector with your thumb in the direction of $\bf{M}$ the loop rotates in the direction your fingers are curled (counter-clockwise). The dipole moment wants to align with the field so $\bf{p_m}$ and $B$ are in the same direction. $\endgroup$ – bremsstrahlung Jan 13 '16 at 20:28
  • $\begingroup$ I don't get it... let's try with Ampere force. The two torques that come from the forces F (in the picture), can be expressed as (Fa/2 sin Alpha), given that F = BIa, the overall torque is (BIa^2 sin Alpha) which is m x B... but that torque doesn't have the direction of M (in the picture). In fact, I think the torque should be directed like that, and not the way it is drawn here... But the cross product says otherwise. $\endgroup$ – Knight1805 Jan 13 '16 at 20:50
  • $\begingroup$ $m \times \bf{B}$ is in the direction of $M$. I have a feeling I'm just misunderstanding what you mean. Could you maybe draw a picture of what you think is correct. Maybe it'll be more clear to me where your confusion is coming from. Also when you say ampere force do you mean Lorentz force? Because the ampere force is the force between two wires, not the force on a wire in a mangetic field. $\endgroup$ – bremsstrahlung Jan 13 '16 at 21:11
  • $\begingroup$ My problem is that such a direction of vector M cannot cause rotation! The axis of rotation of this rectangle is along the vector M in the picture. It divides the loop into two equal parts. It's horizontal. If m x B is the vector of the torque, then the loop should not rotate. My problem is that the orientation of M should not be pointed like this. It should have the same direction as the vectors F. PS. My bad. It is the Lorentz force... when applied to a wire of length a with current I flowing through it, when induction is B is (I a x B) $\endgroup$ – Knight1805 Jan 13 '16 at 21:18

For anyone still having the same problem:

The error here is in fact very fundamental. What I was doing, was taking the torque $M$ from the picture, and comparing it to the force of gravity - not it's appropriate torque.

In fact, what should be done is to find all the torques as vectors, add them up to get a resultant torque. The motion is then described by applying the right hand rule to the overall torque.

It's a bit counter-intuitive but the torque due to force $F$ at length $r$ from the axis of rotation has the direction $r \times F$. Look it up on Wikipedia. Big thanks to the user @bremsstrahlung for clearing up the matter!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.