How to get the magnetic moment for this loop?

[Gourab_chill]

Homework Statement

I've attached the question in the attachments. My question is what is magnetic moment for this loop?

Relevant Equations

μ = I*A

About this figure, the current in the opposite wires are parallel (and not anti-parallel). So, for instance for the first option the torque is zero; but I wanted to know what is the magnetic moment of this loop. Since I rely only on formula I've have no idea how to compute for this one.

 

[TSny]

There is a general formula for the magnetic moment of any current distribution that is covered in upper-division courses. See here. But, you don't need the magnetic moment in order to calculate the torque. Just find the torque by using the force on each section of the loop.

 

[Delta2]

Yes I also think this can be solved just by using the formula $$\vec{F}=I(\vec{l}\times \vec{B})$$
https://en.wikipedia.org/wiki/Lorentz_force#Force_on_a_current-carrying_wire
where $\vec{l}$ is the vector for each segment of the loop pointing in the direction of the current $I$ of that segment.
And then with vectorial addition of the four different forces and four different torques.

 

[Gourab_chill]

There is a general formula for the magnetic moment of any current distribution that is covered in upper-division courses. See here. But, you don't need the magnetic moment in order to calculate the torque. Just find the torque by using the force on each section of the loop.

No as I already told in the explanations section I have found the torque by the the equation @Delta2 has mentioned; is it impossible to calculate the magnetic moment for this loop? The link you have provided contains formula and theories beyond my scope (that's the reason i don't go to Wikipedia:) )

 

[Delta2]

No as I already told in the explanations section I have found the torque by the the equation @Delta2 has mentioned; is it impossible to calculate the magnetic moment for this loop? The link you have provided contains formula and theories beyond my scope (that's the reason i don't go to Wikipedia:) )

Well I believe the article at Wikipedia contains the answer to your question (so it is not beyond your scope).

According to the article the magnetic moment is said to be $$\vec{m}=\frac{1}{2}\iiint \vec{r}\times \vec{J}(\vec{r}) d^3\vec{r}$$ which is a triple (volume) integral and contains the current density vector $\vec{J}$.

However in the case of infinitesimally thin wires I believe it is simplified to the line integral $$\vec{m}=\frac{1}{2}\int_C I(\vec{r})\vec{r}\times d\vec{r}$$ where the domain of integration $C$ is over the circuit or the closed loop and $\vec{r}$ is the position vector and $I(\vec{r})$ is the value of current at position $\vec{r}$.
So in your case $C$ is the square loop, you can break the line integral to 4 parts and do the integration ( i believe the use of cartesian coordinate system is the best here).

 

[TSny]

Consider the loop shown

The loop has two different currents $I_1$ and $I_2$. For such a loop, the general formula for the magnetic moment can be simplified. The result for such a loop depends on the choice of origin, shown as point $O$ above. For your loop, a natural choice of origin would be the center of the square. It can be shown that the general formula reduces to doing the following. Use the origin to construct the two areas $A_1$ and $A_2$ shown below

We will use the sign convention that if the current flows counter-clockwise around the origin, then the contribution to the magnetic moment is positive. This will be the case for $I_1$. $I_2$ is clockwise around the origin and will contribute negatively. The contribution of either current to the magnetic moment is just the current times the associated area.

So, the magnetic moment of the total loop, relative to the origin shown, is given by $I_1A_1 - I_2 A_2$.

 

[Delta2]

@TSny I have done the calculation of the line integral presented at my post #5 (with origin at the left down corner of the square loop and in cartesian coordinates) and I get a result that matches your result in post #6. Specifically I get (L the side of the square loop) $\frac{1}{2}L^2(I_1-I_2)\hat z$ and with $\frac{1}{2}L^2$ being the area of the square above or below the diagonal I believe this matches your result.

But I believe the line integral (as well as the volume integral) are independent of the origin of the coordinate system we choose to do the calculations.

 

[TSny]

@TSny I have done the calculation of the line integral presented at my post #5 (with origin at the left down corner of the square loop) and I get a result that matches your result in post #6.

Good!

But I believe the line integral (as well as the volume integral) are independent of the origin of the coordinate system we choose to do the calculations.

For the usual case where you have a loop where the current has the same magnitude at every point and the current flows in only one direction around the loop, the magnetic moment $\vec m$ will be independent of the choice of origin. However, for the square loop in this problem, $\vec m$ will depend on the location of the origin.

Consider your formula from post #5: $\,\,\, \vec m = \frac{1}{2}\int_C I \,\vec r \times d\vec s$.
Here, $d \vec s$ is an element of displacement along the loop $C$ in the direction of the current, and $\vec r$ is the location of the element relative to the origin. $I$ is the current at the location of the element $d\vec s$.

Shift to a new origin and let $\vec r'$ denote positions of elements of the loop relative to the new origin. So, $\vec r' = \vec r + \vec r_0$, where $\vec r_0$ is a constant vector. Then, relative to the new origin

$\vec m' = \frac{1}{2}\int_C I \, \vec r' \times d\vec s = \frac{1}{2}\int_C I\, \vec r \times d\vec s +\frac{1}{2}\int_C I \,\vec r_0 \times d\vec s = \vec m + \frac{1}{2} \vec r_0 \times \int_C I \, d\vec s$

So, $\vec m' = \vec m$ only if the expression $\vec r_0 \times \int_C I \, d\vec s = 0$. For the typical current loop, $\int_C I \, d\vec s = I \int_C d\vec s = 0$. But, for a loop with junctions, such as the square loop in this problem, $\vec r_0 \times \int_C I \, d\vec s \neq 0$ in general.

 

[Delta2]

Well thanks @TSny but there is something that bugs me. In your proof you use $d\vec{s}$ and omit the dependency of I on $\vec{r}$ I guess you mean $I(\vec{s})$ there.

However if we use my formula directly as it is at post #5 we get that
$$\vec{m}'=\frac{1}{2}\int_C I(\vec{r'})\vec{r'}\times d\vec{r'}$$ and though $d\vec{r'}=d\vec{r}$ , it would however be $I(\vec{r'})\neq I(\vec{r})$ and the proof cannot continue along the same route.

So is my formula wrong exactly as it is written at post #5?

 

[TSny]

So is my formula wrong exactly as it is written at post #5?

Your formula looks good to me. When switching to a new origin, you have to replace $I(\vec r)$ by a different function $I'(\vec r')$. If $\vec r$ and $\vec r'$ point to the same current element $d\vec s$, then $I(\vec r) = I'(\vec r')$. But, as you say, $I(\vec r) \neq I(\vec r')$.

$\vec m'$ would be written as $$\vec m' = \frac{1}{2} \int_C I'(\vec r') \vec r' \times d\vec r'$$ I don't see any difficulty here.