^{In Prob. 6.4, you calculated the force on a dipole by "brute force." Here's a more elegant approach. First write$\mathbf{B}\left(\mathbf{r}\right)$as a Taylor expansion about the center of the loop:}

^{,$\mathbf{B}\left(\mathbf{r}\right)\cong \mathbf{B}\left({\mathbf{r}}_0\right)+[(\mathbf{r}-{\mathbf{r}}_0)\cdot {\nabla }_0]\mathbf{B}\left({\mathbf{r}}_0\right)$}

^{Where${\mathbf{r}}_0$the position of the dipole and ${\nabla }_0$is denotes differentiation with respect to${\mathbf{r}}_0$. Put this into the Lorentz force law (Eq. 5.16) to obtain}

^{.$\mathbf{F}=I\oint d\mathbf{I}\times \left[(\mathbf{r}\cdot {\nabla }_0)\mathbf{B}\left({\mathbf{r}}_0\right)\right]$}

^{Or, numbering the Cartesian coordinates from 1 to 3:}

^{${\mathbf{F}}_i=I\sum _{\mathbf{j}\mathbf{,}\mathbf{k}\mathbf{,}\mathbf{l}=\mathbf{1}}^3{\epsilon }_{ijk}\{\oint {\mathbf{r}}_{l}d{l}_j\}\left[{\nabla }_{\mathbf{0}l}{\mathbf{B}}_k\left({\mathbf{r}}_0\right)\right]$,}

^{Where ${\epsilon }_{\mathrm{ijk}}$ is the Levi-Civita symbol ($+\mathbf{1}$if$\mathbf{ijk}=\mathbf{123}\mathbf{,}\mathbf{231}$, or$\mathbf{312}$; $-\mathbf{1}$if$\mathbf{ijk}=\mathbf{132}$, $\mathbf{213}$, or $\mathbf{321}\mathbf{;}\mathbf{0}$otherwise), in terms of which the cross-product can be written ${(\mathbf{A}\times \mathbf{B})}_i={\sum }_{\mathbf{j}\mathbf{,}\mathbf{k}=\mathbf{1}}^3{\epsilon }_{\mathrm{ijk}}{\mathbf{A}}_j{\mathbf{B}}_k$. Use Eq. 1.108 to evaluate the integral. Note that}

^{$\sum _{\mathbf{j}=\mathbf{1}}^3{\epsilon }_{\mathrm{ijk}}{\epsilon }_{\mathrm{ljm}}={\delta }_{\mathrm{il}}{\delta }_{\mathrm{km}}-{\delta }_{\mathrm{im}}{\delta }_{\mathrm{kl}}$}

^{Whereoil is the Kronecker delta (Prob. 3.52).}

Consider the as a Taylor expansion about the center of the loop:

Write the formula of Lorentz force on a dipole of vector notation.

$\overrightarrow{\mathbf{F}}=\mathbf{I}[{\nabla }_0(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{B}}\left({\overrightarrow{\mathbf{r}}}_0\right))-\overrightarrow{\mathbf{a}}({\nabla }_0\cdot \overrightarrow{\mathbf{B}}\left({\overrightarrow{\mathbf{r}}}_0\right))]$…… (1)

Here,${\nabla }_0$ is zero coordinates, $\mathbf{B}\left({\mathbf{r}}_0\right)$ is Taylor expansion and is current.

Determine the magnetic field is:

$\overrightarrow{B}\left(\overrightarrow{r}\right)\approx \overrightarrow{B}\left({\overrightarrow{r}}_0\right)+[(\overrightarrow{r}-{\overrightarrow{r}}_0)\cdot {\nabla }_0]\overrightarrow{B}\left({\overrightarrow{r}}_0\right)$

Here, ${\nabla }_0$ means it operates on “zeroed” coordinates. Put this into the Lorentz force law:

$\begin{array}{c}\overrightarrow{F}=\oint I(d\overrightarrow{I}\times \overrightarrow{B})\\ =I\oint d\overrightarrow{I}\times [\overrightarrow{B}\left({\overrightarrow{r}}_0\right)+[(\overrightarrow{r}-{\overrightarrow{r}}_0)\cdot {\nabla }_0]\overrightarrow{B}\left({\overrightarrow{r}}_0\right)]\end{array}$

In order to allow "zeroed" terms to leave the integral, we integrate over the "un-zeroed" coordinates. The integral of dl over a closed loop is zero, hence we can utilise this knowledge to obtain:

$\overrightarrow{F}=I\oint d\overrightarrow{I}\times \left[(\overrightarrow{r}\cdot {\nabla }_0)\overrightarrow{B}\left({\overrightarrow{r}}_0\right)\right]$

The integrand in the indicial notation is:

${\epsilon }_{ijk}d{l}_j\left(r_n{\partial }_{0n}{B}_{0k}\right)$

The Einstein summation convention will be used extensively. If an index is used twice in one phrase, it means that the values have been summed (in this case 1,2 and 3). The notation's integral is as follows:

$\begin{array}{c}{F}_i=I\oint {\epsilon }_{ijk}d{l}_j\left(r_n{\partial }_{0n}{B}_{0k}\right)\\ =I{\epsilon }_{ijk}(\oint d{l}_j{r}_n){\partial }_{0n}{B}_{0k}\end{array}$

Now let's concentrate on the phrase in brackets. To what does it equate? Well, 1.108 indicates:

$\oint (\overrightarrow{c}\cdot \overrightarrow{r})d\overrightarrow{l}=\overrightarrow{a}\times \overrightarrow{c}$

In indicial notation this is:

$\oint c_i{r}_{i}d{l}_n={\epsilon }_{nij}{a}_1{c}_j$

We chose $r_n={\delta }_{nl}rl$to obtain the phrase in brackets above, resulting in:

$\begin{array}{c}\oint d{l}_j{r}_n=\oint {\delta }_{nl}rld{l}_j\\ ={\epsilon }_j{l}_P{a}_l{\delta }_{Pn}\\ ={\epsilon }_j\ln a_l\end{array}$

Note: It doesn't matter what the repeated, or "dummy," indices are termed as long as they don't repeat more than twice and the live, or "liver," indices are the same on both sides of the equal sign.

Now:

$\begin{array}{c}{F}_i=I{\epsilon }_{ijk}{\epsilon }_{j\ln }{a}_l{\partial }_{0n}{B}_{0k}\\ =I{\epsilon }_{ijk}{\epsilon }_{j\ln }{a}_l{\partial }_{0n}{B}_{0k}\end{array}$

We use the $\epsilon -\delta$rule:

$\begin{array}{c}{\epsilon }_{ijk}{\epsilon }_{j\ln }=-{\epsilon }_{jik}{\epsilon }_{jln}\\ =-({\delta }_{il}{\delta }_{kn}-{\delta }_{in}{\delta }_{kl})\end{array}$

Now force indices:

$\begin{array}{c}{F}_i=-I({\delta }_{il}{\delta }_{kn}-{\delta }_{in}{\delta }_{kl})a_l{\partial }_{0n}{B}_{0k}\\ =-(a_i{\partial }_{0k}{B}_{0k}-a_k{\partial }_{0i}{B}_{0k})\\ =I({\partial }_{0i}\left(a_k{B}_{0k}\right)-a_i{\partial }_{0k}{B}_{0k})\end{array}$

Since $a_k$ is a constant (it is not "zeroed" in this situation), it might enter the integral. Returning to vector notation, we get the following:

Determine the Lorentz force on a dipole of vector notation.

Substitute $\overrightarrow{m}$ for $I\overrightarrow{a}$ into equation (1).

$\begin{array}{c}\overrightarrow{F}=I{\nabla }_0(\overrightarrow{a}\cdot \overrightarrow{B}\left({\overrightarrow{r}}_0\right))\\ ={\nabla }_0(\overrightarrow{m}\cdot \overrightarrow{B}\left({\overrightarrow{r}}_0\right))\end{array}$

Therefore, the value of Lorentz force on a dipole of vector notation is .

$\overrightarrow{F}={\nabla }_0(\overrightarrow{m}\cdot \overrightarrow{B}\left(r_0\right))$