Can we derive the magnetic dipole moment of a current loop from the Lorentz force on it?

Question

I would like to derive $\boldsymbol{\mu}$ for a current loop where $\boldsymbol{\mu}$ is defined by the relationship between torque on the current loop resulting from it being in a magnetic field, $$\boldsymbol{\tau}=\boldsymbol{\mu}\times\textbf{B}$$ What I have so far is $$d\textbf{F}=Id \textbf{l}\times\textbf{B}$$ from the Lorentz force law. Substituting this into the torque equation, we have $$d\boldsymbol{\tau}=I\textbf{r}\times(d \textbf{l}\times\textbf{B})$$ Im not sure how I should compute this integral though, does anyone know if this calculation has been done before?

Answer 1

Contribution to net torque due to element $d\mathbf r$ is, as you wrote $$d\boldsymbol{\tau}=I\mathbf{r}\times(d \mathbf{r}\times\mathbf{B}).$$ Instead of $d\mathbf l$, we're using symbol $d\mathbf r$ as the vector $d\mathbf l$ is an infinitesimal change in $\mathbf r$.

This can be rewritten using the relation

$$ \mathbf A \times (\mathbf B\times \mathbf C) = \mathbf B(\mathbf A\cdot \mathbf C) - \mathbf C(\mathbf A\cdot \mathbf B)~~~\tag{*} $$ so we have

$$ d\boldsymbol{\tau} = Id\mathbf r(\mathbf r \cdot \mathbf B) - \mathbf B (\mathbf r\cdot d\mathbf r). $$

Integrating over the closed path $\gamma$ where the current flows, the second term gives zero contribution, because it is proportional to $d(r^2)$, and $r^2$ is a function of coordinates only, so it comes to the initial value after a full round trip, so integrating its differential produces zero. So we have

$$ \boldsymbol{\tau} = \oint_\gamma Id\mathbf r(\mathbf r \cdot \mathbf B). $$

We can write this in cartesian coordinates: $$ \boldsymbol{\tau}_i = \oint_\gamma Idr_i\bigg(\sum_s r_s B_s\bigg). $$

Now we make an unintuituive rewrite: we write the integrand as sum of two halves: $$ \boldsymbol{\tau}_i = \oint_\gamma \frac{1}{2}Idr_i\bigg(\sum_s r_s B_s\bigg) + \frac{1}{2}Idr_i\bigg(\sum_s r_s B_s\bigg). $$

Now we rewrite the second integrand using the relation

$$ dr_i r_s B_s = d(r_i r_s B_s)-r_i dr_s B_s; $$ this holds because $B_s$ is assumed uniform in the whole region, so $dB_s=0$. Thus the second integrand is

$$ \frac{1}{2}I\bigg(\sum_s dr_i r_s B_s\bigg) = \frac{1}{2}I\bigg(\sum_s d(r_i r_s B_s) - \sum_s r_i dr_s B_s\bigg). $$

Writing the integral using this expression, we get

$$ \boldsymbol{\tau}_i = \oint_\gamma \frac{1}{2}Idr_i\bigg(\sum_s r_s B_s\bigg) + \frac{1}{2}I\bigg(\sum_s d(r_i r_s B_s) - \sum_s r_i dr_s B_s\bigg). $$

The terms proportional to $d(r_i r_sB_s)$ integrate to zero, so we can drop those terms and then we obtain $$ \boldsymbol{\tau}_i = \oint_\gamma \frac{1}{2}Idr_i\bigg(\sum_s r_s B_s\bigg) - \frac{1}{2}Ir_i \bigg( \sum_s dr_s B_s\bigg), $$ or in vector notation $$ \boldsymbol{\tau} = \oint_\gamma \frac{1}{2}Id\mathbf r (\mathbf r\cdot \mathbf B) - \frac{1}{2}I \mathbf r(d\mathbf r\cdot \mathbf B). $$

Using the formula (*) in reverse, we can see that

$$ \boldsymbol{\tau} = \bigg( \oint_\gamma \frac{1}{2}I\mathbf r \times d\mathbf r\bigg) \times \mathbf B. $$

The last integral is current times the surface area vector of the closed loop, which is, by definition, the magnetic moment of the current loop:

$$ \boldsymbol{\mu} = I \oint_\gamma \frac{1}{2}\mathbf r \times d\mathbf r . $$