91影视

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

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

$\stackrel{\mathbf{鈫�}}{\mathbf{F}}=\mathbf{I}[{鈭�}_0(\stackrel{\mathbf{鈫�}}{\mathbf{a}}鈰�\stackrel{\mathbf{鈫�}}{\mathbf{B}}\left({\stackrel{\mathbf{鈫�}}{\mathbf{r}}}_0\right))鈭�\stackrel{\mathbf{鈫�}}{\mathbf{a}}({鈭�}_0鈰�\stackrel{\mathbf{鈫�}}{\mathbf{B}}\left({\stackrel{\mathbf{鈫�}}{\mathbf{r}}}_0\right))]$鈥︹�� (1)

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

Determine the magnetic field is:

$\stackrel{鈫�}{B}\left(\stackrel{鈫�}{r}\right)鈮�\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right)+[(\stackrel{鈫�}{r}鈭�{\stackrel{鈫�}{r}}_0)鈰�{鈭�}_0]\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right)$

Here, ${鈭�}_0$ means it operates on 鈥渮eroed鈥� coordinates. Put this into the Lorentz force law:

$\begin{array}{c}\stackrel{鈫�}{F}=鈭�I(d\stackrel{鈫�}{I}脳\stackrel{鈫�}{B})\\ =I鈭�d\stackrel{鈫�}{I}脳[\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right)+[(\stackrel{鈫�}{r}鈭�{\stackrel{鈫�}{r}}_0)鈰�{鈭�}_0]\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right)]\end{array}$

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:

$\stackrel{鈫�}{F}=I鈭�d\stackrel{鈫�}{I}脳\left[(\stackrel{鈫�}{r}鈰�{鈭�}_0)\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right)\right]$

The integrand in the indicial notation is:

${蔚}_{ijk}d{l}_j\left(r_n{鈭�}_{0n}{B}_{0k}\right)$

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:

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

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

$鈭�(\stackrel{鈫�}{c}鈰�\stackrel{鈫�}{r})d\stackrel{鈫�}{l}=\stackrel{鈫�}{a}脳\stackrel{鈫�}{c}$

In indicial notation this is:

$鈭�c_i{r}_{i}d{l}_n={蔚}_{nij}{a}_1{c}_j$

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

$\begin{array}{c}鈭�d{l}_j{r}_n=鈭�{未}_{nl}rld{l}_j\\ ={蔚}_j{l}_P{a}_l{未}_{Pn}\\ ={蔚}_j\ln a_l\end{array}$

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:

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

We use the $蔚鈭�未$rule:

$\begin{array}{c}{蔚}_{ijk}{蔚}_{j\ln }=鈭�{蔚}_{jik}{蔚}_{jln}\\ =鈭�({未}_{il}{未}_{kn}鈭�{未}_{in}{未}_{kl})\end{array}$

Now force indices:

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

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 $\stackrel{鈫�}{m}$ for $I\stackrel{鈫�}{a}$ into equation (1).

$\begin{array}{c}\stackrel{鈫�}{F}=I{鈭�}_0(\stackrel{鈫�}{a}鈰�\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right))\\ ={鈭�}_0(\stackrel{鈫�}{m}鈰�\stackrel{鈫�}{B}\left({\stackrel{鈫�}{r}}_0\right))\end{array}$

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

$\stackrel{鈫�}{F}={鈭�}_0(\stackrel{鈫�}{m}鈰�\stackrel{鈫�}{B}\left(r_0\right))$