D.72 Var­i­ous elec­tro­sta­tic de­riva­tions.

This sec­tion gives var­i­ous de­riva­tions for the elec­tro­mag­ne­to­sta­tic so­lu­tions of chap­ter 13.3.

D.72.1 Ex­is­tence of a po­ten­tial

This sub­sec­tion shows that if the curl of the elec­tric field $\skew3\vec{\cal E}$ (or of any other vec­tor field, like the mag­netic one or a force field), is zero, it is mi­nus the gra­di­ent of some po­ten­tial.

That po­ten­tial can be de­fined to be

$$\varphi({\skew0\vec r}) = - \int_{{\skew0\vec r}_0}^{\skew0... ...{\underline{\skew0\vec r}}){ \rm d}{\underline{\skew0\vec r}}$$ (D.47)

where ${\skew0\vec r}_0$ is some ar­bi­trar­ily cho­sen ref­er­ence point. You might think that the value of $\varphi({\skew0\vec r})$ would de­pend on what in­te­gra­tion path you took from the ref­er­ence point to ${\skew0\vec r}$, but the Stokes’ the­o­rem of cal­cu­lus says that the dif­fer­ence be­tween in­te­grals lead­ing to the same path must be zero since $\nabla$ $\times$ $\skew3\vec{\cal E}$ is zero.

Now if you eval­u­ate $\varphi$ at a neigh­bor­ing point ${\skew0\vec r}+{\hat\imath}\partial{x}$ by a path first go­ing to ${\skew0\vec r}$ and from there straight to ${\skew0\vec r}+{\hat\imath}\partial{x}$, the dif­fer­ence in in­te­grals is just the in­te­gral over the fi­nal seg­ment:

$$\varphi({\skew0\vec r}+{\hat\imath}\partial x) - \varphi({\... ...{\underline{\skew0\vec r}}){ \rm d}{\underline{\skew0\vec r}}$$ (D.48)

Di­vid­ing by $\partial{x}$ and then tak­ing the limit $\partial{x}\to0$ shows that mi­nus the $x$-​de­riv­a­tive of $\varphi$ gives the $x$-​com­po­nent of the elec­tric field. The same of course for the other com­po­nents, since the $x$-​di­rec­tion is ar­bi­trary.

Note that if re­gions are mul­ti­ply con­nected, the po­ten­tial may not be quite unique. The most im­por­tant ex­am­ple of that is the mag­netic po­ten­tial of an in­fi­nite straight elec­tric wire. Since the curl of the mag­netic field is nonzero in­side the wire, the path of in­te­gra­tion must stay clear of the wire. It then turns out that the value of the po­ten­tial de­pends on how many times the cho­sen in­te­gra­tion path wraps around the wire. In­deed, the mag­netic po­ten­tial is $\varphi_m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$I\theta$$\raisebox{.5pt}{$/$}$​$2\pi\epsilon_0c^2$. and as you know, an an­gle like $\theta$ is in­de­ter­mi­nate by any in­te­ger mul­ti­ple of $2\pi$.

D.72.2 The Laplace equa­tion

The ho­mo­ge­neous Pois­son equa­tion,

$$\nabla^2 \varphi = 0 %$$ (D.49)

for some un­known func­tion $\varphi$ is called the Laplace equa­tion. It is very im­por­tant in many ar­eas of physics and en­gi­neer­ing. This note de­rives some of its generic prop­er­ties.

The so-called mean-value prop­erty says that the av­er­age of $\varphi$ over the sur­face of any sphere in which the Laplace equa­tion holds is the value of $\varphi$ at the cen­ter of the sphere. To see why, for con­ve­nience take the cen­ter of the sphere as the ori­gin of a spher­i­cal co­or­di­nate sys­tem. Now

$$\begin{eqnarray*} 0 & = & \int_{\rm sphere} \nabla^2 \varphi { \rm d}^3{\skew0... ...n-7pt\int}\nolimits \varphi\sin\theta { \rm d}\theta{\rm d}\phi \end{eqnarray*}$$

the first equal­ity since $\varphi$ sat­is­fies the Laplace equa­tion, the sec­ond be­cause of the di­ver­gence the­o­rem, the third be­cause the in­te­gral is zero, so a con­stant fac­tor does not make a dif­fer­ence, and the fourth by chang­ing the or­der of in­te­gra­tion and dif­fer­en­ti­a­tion. It fol­lows that the av­er­age of $\varphi$ is the same on all spher­i­cal sur­faces cen­tered around the ori­gin. Since this in­cludes as a lim­it­ing case the ori­gin and the av­er­age of $\varphi$ over the sin­gle point at the ori­gin is just $\varphi$ at the ori­gin, the mean value prop­erty fol­lows.

The so called max­i­mum-min­i­mum prin­ci­ple says that ei­ther $\varphi$ is con­stant every­where or its max­i­mum and min­i­mum are on a bound­ary or at in­fin­ity. The rea­son is the mean-value prop­erty above. Sup­pose there is an ab­solute max­i­mum in the in­te­rior of the re­gion in which the Laplace equa­tion ap­plies. En­close the max­i­mum by a small sphere. Since the val­ues of $\varphi$ would be less than the max­i­mum on the sur­face of the sphere, the av­er­age value on the sur­face must be less than the max­i­mum too. But the mean value the­o­rem says it must be the same. The only way around that is if $\varphi$ is com­pletely con­stant in the sphere, but then the max­i­mum is not a true max­i­mum. And then you can start sphere-hop­ping to show that $\varphi$ is con­stant every­where. Min­ima go the same way.

The only so­lu­tion of the Laplace equa­tion in all of space that is zero at in­fin­ity is zero every­where. In more gen­eral re­gions, as long as the so­lu­tion is zero on all bound­aries, in­clud­ing in­fin­ity where rel­e­vant, then the so­lu­tion is zero every­where. The rea­son is the max­i­mum-min­i­mum prin­ci­ple: if there was a point where the so­lu­tion was pos­i­tive/neg­a­tive, then there would have to be an in­te­rior max­i­mum/min­i­mum some­where.

The so­lu­tion of the Laplace equa­tion for given bound­ary val­ues is unique. The rea­son is that the dif­fer­ence be­tween any two so­lu­tions must sat­isfy the Laplace equa­tion with zero bound­ary val­ues, hence must be zero.

D.72.3 Egg-shaped di­pole field lines

The egg shape of the ideal di­pole field lines can be found by as­sum­ing that the di­pole is di­rected along the $z$-​axis. Then the field lines in the $xz$-​plane sat­isfy

$$\frac{{\rm d}z}{{\rm d}x}= \frac{{\cal E}_z}{{\cal E}_x} = \frac{2z^2-x^2}{3zx}$$

Change to a new vari­able $u$ by re­plac­ing $z$ by $xu$ to get:

$$x \frac{{\rm d}u}{{\rm d}x}= - \frac{1+u^2}{3u} \quad\Long... ...d \int \frac{3u{ \rm d}u}{1+u^2} = - \int \frac{{\rm d}x}{x}$$

In­te­grat­ing and re­plac­ing $u$ again by $z$$\raisebox{.5pt}{$/$}$​$x$ gives

$$(x^2+z^2)^{3/2} = C x^2$$

where $C$ rep­re­sents the in­te­gra­tion con­stant from the in­te­gra­tion. Near the ori­gin, $x$ $\vphantom0\raisebox{1.5pt}{$\sim$}$ $z^{3/2}$$\raisebox{.5pt}{$/$}$​$C$; there­fore the field line has in­fi­nite cur­va­ture at the ori­gin, ex­plain­ing the pro­nounced egg shape. Rewrit­ten in spher­i­cal co­or­di­nates, the field lines are given by $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $C\sin^2\theta$ and $\phi$ con­stant, and that is also valid out­side the $xz$-​plane.

D.72.4 Ideal charge di­pole delta func­tion

Next is the delta func­tion in the elec­tric field gen­er­ated by a charge dis­tri­b­u­tion that is con­tracted to an ideal di­pole. To find the pre­cise delta func­tion, the elec­tric field can be in­te­grated over a small sphere, but still large enough that on its sur­face the ideal di­pole po­ten­tial is valid. The in­te­gral will give the strength of the delta func­tion. Since the elec­tric field is mi­nus the gra­di­ent of the po­ten­tial, an ar­bi­trary com­po­nent ${\cal E}_i$ in­te­grates to

$$\int_{\rm sphere} {\cal E}_i { \rm d}^3{\skew0\vec r} = -... ...\vec r} = - \int_{\rm sphere surface} \varphi n_i { \rm d}A$$

where ${\hat\imath}_i$ is the unit vec­tor in the $i$-​di­rec­tion and the di­ver­gence the­o­rem of cal­cu­lus was used to con­vert the in­te­gral to an in­te­gral over the sur­face area $A$ of the sphere. Not­ing that the vec­tor ${\vec n}$ nor­mal to the sur­face of the sphere equals ${\skew0\vec r}$$\raisebox{.5pt}{$/$}$​$r$, and that the po­ten­tial is the ideal di­pole one, you get

$$\int_{\rm sphere} {\cal E}_i { \rm d}^3{\skew0\vec r}= - \... ...rac{\vec\wp\cdot{\skew0\vec r}}{r^3} \frac{r_i}{r} { \rm d}A$$

For sim­plic­ity, take the $z$-​axis along the di­pole mo­ment; then $\vec\wp\cdot{\skew0\vec r}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\wp}z$. For the $x$-​com­po­nent ${\cal E}_x$, $r_i$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$ so that the in­te­grand is pro­por­tional to $xz$, and that in­te­grates to zero over the sur­face of the sphere be­cause the neg­a­tive $x$-​val­ues can­cel the pos­i­tive ones at the same $z$. The same for the $y$-​com­po­nent of the field, so only the $z$-​com­po­nent, or more gen­er­ally, the com­po­nent in the same di­rec­tion as $\vec\wp$, has a delta func­tion. For ${\cal E}_z$, you are in­te­grat­ing $z^2$, and by sym­me­try that is the same as in­te­grat­ing $x^2$ or $y^2$, so it is the same as in­te­grat­ing $\frac13r^2$. Since the sur­face of the sphere equals $4{\pi}r^2$, the delta func­tion in­cluded in the ex­pres­sion for the field of a di­pole as listed in ta­ble 13.2 is ob­tained.

D.72.5 In­te­grals of the cur­rent den­sity

In sub­se­quent de­riva­tions, var­i­ous in­te­grals of the cur­rent den­sity $\vec\jmath$ are needed. In all cases it is as­sumed that the cur­rent den­sity van­ishes strongly out­side some re­gion. Of course, nor­mally an elec­tric mo­tor or elec­tro­mag­net has elec­tri­cal leads go­ing to­wards and away from of it; it is as­sumed that these are stranded so tightly to­gether that their net ef­fect can be ig­nored.

Con­sider an in­te­gral like $\int{r_i}^mr_{\overline{\imath}}^nj_i{ \rm d}^3{\skew0\vec r}$ where $j_i$ is any com­po­nent $j_1$, $j_2$, or $j_3$ of the cur­rent den­sity, ${\overline{\imath}}$ is the in­dex fol­low­ing $i$ in the se­quence $\ldots123123\ldots$, $m$ and $n$ are non­neg­a­tive in­te­gers, and the in­te­gra­tion is over all of space. By in­te­gra­tion by parts in the $i$-​di­rec­tion, and us­ing the fact that the cur­rent den­si­ties van­ish at in­fin­ity,

$$\int r_i^m r_{\overline{\imath}}^n j_i { \rm d}^3{\skew0\v... ...n \frac{\partial j_i}{\partial r_i} { \rm d}^3{\skew0\vec r}$$

Now use the fact that the di­ver­gence of the cur­rent den­sity is zero since the charge den­sity is con­stant for elec­tro­mag­ne­to­sta­tic so­lu­tions:

$$\int r_i^m r_{\overline{\imath}}^n j_i { \rm d}^3{\skew0\v... ...l r_{\overline{\overline{\imath}}}} { \rm d}^3{\skew0\vec r}$$

where ${\overline{\overline{\imath}}}$ is the in­dex pre­ced­ing $i$ in the se­quence $\ldots123123\ldots$. The fi­nal in­te­gral can be in­te­grated in the ${\overline{\overline{\imath}}}$-​di­rec­tion and is then seen to be zero be­cause $\vec\jmath$ van­ishes at in­fin­ity.

The first in­te­gral in the right hand side can be in­te­grated by parts in the ${\overline{\imath}}$-​di­rec­tion to give the fi­nal re­sult:

$$\int r_i^m r_{\overline{\imath}}^n j_i { \rm d}^3{\skew0\v... ...imath}}^{n-1} j_{\overline{\imath}}{ \rm d}^3{\skew0\vec r} %$$ (D.50)

It fol­lows from this equa­tion with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 that

$$\int r_i j_{\overline{\imath}}{ \rm d}^3{\skew0\vec r}= - ... ...int {\skew0\vec r}\times\vec\jmath { \rm d}^3{\skew0\vec r} %$$ (D.51)

with $\vec\mu$ the cur­rent dis­tri­b­u­tion’s di­pole mo­ment. In these ex­pres­sions, you can swap in­dices as

$$(i,{\overline{\imath}},{\overline{\overline{\imath}}}) \to ... ...}) \to ({\overline{\overline{\imath}}},i,{\overline{\imath}})$$

be­cause only the rel­a­tive or­der­ing of the in­dices in the se­quence $\ldots123123\ldots$ is rel­e­vant.

In quan­tum ap­pli­ca­tions, it is of­ten nec­es­sary to re­late the di­pole mo­ment to the an­gu­lar mo­men­tum of the cur­rent car­ri­ers. Since the cur­rent den­sity is the charge per unit vol­ume times its ve­loc­ity, you get the lin­ear mo­men­tum per unit vol­ume by mul­ti­ply­ing by the ra­tio $m_{\rm {c}}$$\raisebox{.5pt}{$/$}$​$q_{\rm {c}}$ of cur­rent car­rier mass over charge. Then the an­gu­lar mo­men­tum is

$$\vec L = \int {\skew0\vec r}\times \frac{m_{\rm c}}{q_{\rm ... ...\rm d}^3{\skew0\vec r} = \frac{2m_{\rm c}}{q_{\rm c}} \vec\mu$$

D.72.6 Lorentz forces on a cur­rent dis­tri­b­u­tion

Next is the de­riva­tion of the Lorentz forces on a given cur­rent dis­tri­b­u­tion $\vec\jmath$ in a con­stant ex­ter­nal mag­netic field $\skew2\vec{\cal B}_{\rm {ext}}$. The Lorentz force law says that the force $\vec{F}$ on a charge $q$ mov­ing with speed $\vec{v}$ equals

$$\vec F = q \vec v \times \skew2\vec{\cal B}_{\rm ext}$$

In terms of a cur­rent dis­tri­b­u­tion, the mov­ing charge per unit vol­ume times its ve­loc­ity is the cur­rent den­sity, so the force on a vol­ume el­e­ment ${\rm d}^3{\skew0\vec r}$ is:

$${\rm d}\vec F = \vec\jmath \times \skew2\vec{\cal B}_{\rm ext} { \rm d}^3{\skew0\vec r}$$

The net force on the cur­rent dis­tri­b­u­tion is there­fore zero, be­cause ac­cord­ing to (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the in­te­grals of the com­po­nents of the cur­rent dis­tri­b­u­tion are zero.

The mo­ment is not zero, how­ever. It is given by

$$\vec M = \int {\skew0\vec r}\times \left(\vec\jmath\times \skew2\vec{\cal B}_{\rm ext}\right) { \rm d}^3{\skew0\vec r}$$

Ac­cord­ing to the vec­to­r­ial triple prod­uct rule, that is

$$\vec M = \int \left({\skew0\vec r}\cdot \skew2\vec{\cal B}... ...\right) \skew2\vec{\cal B}_{\rm ext} { \rm d}^3{\skew0\vec r}$$

The sec­ond in­te­gral is zero be­cause of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. What is left is can be writ­ten in in­dex no­ta­tion as

$$M_i = \int r_i {\cal B}_{\rm ext,i} j_i { \rm d}^3{\skew0... ...,{\overline{\overline{\imath}}}} j_i { \rm d}^3{\skew0\vec r}$$

The first of the three in­te­grals is zero be­cause of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The other two can be rewrit­ten us­ing (D.51):

$$M_i = - \mu_{\overline{\overline{\imath}}}{\cal B}_{\rm ex... ...line{\imath}}{\cal B}_{\rm ext,{\overline{\overline{\imath}}}}$$

and in vec­tor no­ta­tion that reads

$$\vec M = \vec \mu \times \skew2\vec{\cal B}_{\rm ext}$$

When the (frozen) cur­rent dis­tri­b­u­tion is slowly ro­tated around the axis aligned with the mo­ment vec­tor, the work done is

$$- M { \rm d}\alpha = - \mu {\cal B}_{\rm ext} \sin\alpha { \rm d}\alpha = { \rm d}(\mu {\cal B}_{\rm ext} \cos\alpha)$$

where $\alpha$ is the an­gle be­tween $\vec\mu$ and $\skew2\vec{\cal B}_{\rm {ext}}$. By in­te­gra­tion, it fol­lows that the work done cor­re­sponds to a change in en­ergy for an en­ergy given by

$$E_{\rm ext} = - \vec\mu \cdot \skew2\vec{\cal B}_{\rm ext}$$

D.72.7 Field of a cur­rent di­pole

A cur­rent den­sity $\vec\jmath$ cre­ates a mag­netic field be­cause of Maxwell’s sec­ond and fourth equa­tions for the di­ver­gence and curl of the mag­netic field:

$$\nabla \cdot \skew2\vec{\cal B}= 0 \qquad \nabla \times \skew2\vec{\cal B}= \frac{1}{\epsilon_0c^2} \vec\jmath$$

where $\skew2\vec{\cal B}$ van­ishes at in­fin­ity as­sum­ing there is no ad­di­tional am­bi­ent mag­netic field.

A mag­netic vec­tor po­ten­tial $\skew3\vec A$ will now be de­fined as the so­lu­tion of the Pois­son equa­tion

$$\nabla^2 \skew3\vec A= - \frac{1}{\epsilon_0c^2} \vec\jmath$$

that van­ishes at in­fin­ity. Tak­ing the di­ver­gence of this equa­tion shows that the di­ver­gence of the vec­tor po­ten­tial sat­is­fies a ho­mo­ge­neous Pois­son equa­tion, be­cause the di­ver­gence of the cur­rent den­sity is zero, with zero bound­ary con­di­tions at in­fin­ity. There­fore the di­ver­gence of the vec­tor po­ten­tial is zero. It then fol­lows that

$$\skew2\vec{\cal B}= \nabla \times \skew3\vec A$$

be­cause it sat­is­fies the equa­tions for $\skew2\vec{\cal B}$: the di­ver­gence of any curl is zero, and the curl of the curl of the vec­tor po­ten­tial is ac­cord­ing to the vec­to­r­ial triple prod­uct its Lapla­cian, hence the cor­rect curl of the mag­netic field.

You might of course won­der whether there might not be more than one mag­netic field that has the given di­ver­gence and curl and is zero at in­fin­ity. The an­swer is no. The dif­fer­ence be­tween any two such fields must have zero di­ver­gence and curl. There­fore the curl of the curl of the dif­fer­ence is zero too, and the vec­to­r­ial triple prod­uct shows that equal to mi­nus the Lapla­cian of the dif­fer­ence. If the Lapla­cian of the dif­fer­ence is zero, then the dif­fer­ence is zero, since the dif­fer­ence is zero at in­fin­ity (sub­sec­tion 2). So the so­lu­tions must be the same.

Since the in­te­grals of the cur­rent den­sity are zero, (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, the as­ymp­totic ex­pan­sion (13.31) of the Green’s func­tion in­te­gral shows that at large dis­tances, the com­po­nents of $\skew3\vec A$ be­have as a di­pole po­ten­tial. Specif­i­cally,

$$A_i \sim \frac{1}{4\pi\epsilon_0c^2r^3} \sum_{{\underline ... ...ine r}_{\underline i}j_i { \rm d}^3{\underline{\skew0\vec r}}$$

Now the term ${\underline i}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $i$ in the sum does not give a con­tri­bu­tion, be­cause of (D.50) with $m$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. The other two terms are

$$A_i \sim \frac{1}{4\pi\epsilon_0c^2r^3} \left[ r_{\overli... ...ne{\imath}}}j_i { \rm d}^3{\underline{\skew0\vec r}} \right]$$

with ${\overline{\imath}}$ fol­low­ing $i$ in the se­quence $\ldots123123\ldots$ and ${\overline{\overline{\imath}}}$ pre­ced­ing it. These two in­te­grals can be rewrit­ten us­ing (D.51) to give

$$A_i \sim - \frac{1}{4\pi\epsilon_0c^2r^3} \left[ r_{\over... ...{\overline{\overline{\imath}}}\mu_{\overline{\imath}} \right]$$

Note that the ex­pres­sion be­tween brack­ets is just the $i$-​th com­po­nent of ${\skew0\vec r}$ $\times$ $\vec\mu$.

The mag­netic field is the curl of $\skew3\vec A$, so

$${\cal B}_i = \frac{\partial A_{\overline{\overline{\imath}}... ...\overline{\imath}}}{\partial r_{\overline{\overline{\imath}}}}$$

and sub­sti­tut­ing in for the vec­tor po­ten­tial from above, dif­fer­en­ti­at­ing, and clean­ing up pro­duces

$${\cal B}_i = \frac{3(\vec\mu\cdot{\skew0\vec r}){\skew0\vec r}- \vec\mu r^2}{4\pi\epsilon_0c^2r^5}$$

This is the same as­ymp­totic field as a charge di­pole with strength $\vec\mu$ would have.

How­ever, for an ideal cur­rent di­pole, the delta func­tion at the ori­gin will be dif­fer­ent than that de­rived for a charge di­pole in the first sub­sec­tion. In­te­grate the mag­netic field over a sphere large enough that on its sur­face, the as­ymp­totic field is ac­cu­rate:

$$\int {\cal B}_i { \rm d}^3{\skew0\vec r}= \int \frac{\par... ...al r_{\overline{\overline{\imath}}}} { \rm d}^3{\skew0\vec r}$$

Us­ing the di­ver­gence the­o­rem, the right hand side be­comes an in­te­gral over the sur­face of the sphere:

$$\int {\cal B}_i { \rm d}^3{\skew0\vec r}= \int A_{\overli... ...{\imath}}\frac{r_{\overline{\overline{\imath}}}}{r} { \rm d}A$$

Sub­sti­tut­ing in the as­ymp­totic ex­pres­sion for $A_i$ above,

$$\int {\cal B}_i { \rm d}^3{\skew0\vec r}= - \frac{1}{4\pi\... ...\imath}}}) r_{\overline{\overline{\imath}}}{ \rm d}A \right]$$

The in­te­grals of $r_{i}r_{{\overline{\imath}}}$ and $r_{i}r_{{\overline{\overline{\imath}}}}$ are zero, for one be­cause the in­te­grand is odd in $r_i$. The in­te­grals of $r_{{\overline{\imath}}}r_{\overline{\imath}}$ and $r_{{\overline{\overline{\imath}}}}r_{{\overline{\overline{\imath}}}}$ are each one third of the in­te­gral of $r^2$ be­cause of sym­me­try. So, not­ing that the sur­face area $A$ of the spher­i­cal sur­face is $4{\pi}r^2$,

$$\int {\cal B}_i { \rm d}^3{\skew0\vec r}= \frac{2}{3\epsilon_0c^2} \mu_i$$

That gives the strength of the delta func­tion for an ideal cur­rent di­pole.

D.72.8 Biot-Savart law

In the pre­vi­ous sec­tion, it was noted that the mag­netic field of a cur­rent dis­tri­b­u­tion is the curl of a vec­tor po­ten­tial $\skew3\vec A$. This vec­tor po­ten­tial sat­is­fies the Pois­son equa­tion

$$\nabla^2 \skew3\vec A= - \frac{1}{\epsilon_0c^2} \vec\jmath$$

The so­lu­tion for the vec­tor po­ten­tial can be writ­ten ex­plic­itly in terms of the cur­rent den­sity us­ing the Green’s func­tion in­te­gral (13.29):

$$A_i = \frac{1}{4\pi\epsilon_0c^2} \int \frac{1}{\vert{\ske... ...derline{\skew0\vec r}}) { \rm d}^3 {\underline{\skew0\vec r}}$$

The mag­netic field is the curl of $\skew3\vec A$,

$${\cal B}_i = \frac{\partial A_{\overline{\overline{\imath}}... ...\overline{\imath}}}{\partial r_{\overline{\overline{\imath}}}}$$

or sub­sti­tut­ing in and dif­fer­en­ti­at­ing un­der the in­te­gral

$${\cal B}_i = - \frac{1}{4\pi\epsilon_0c^2} \int \frac{r_{\... ...derline{\skew0\vec r}}) { \rm d}^3 {\underline{\skew0\vec r}}$$

In vec­tor no­ta­tion that gives the Biot-Savart law

$$\skew2\vec{\cal B}= - \frac{1}{4\pi\epsilon_0c^2} \int \fr... ...rt^3} \times \vec\jmath { \rm d}^3 {\underline{\skew0\vec r}}$$

Now as­sume that the cur­rent dis­tri­b­u­tion is lim­ited to one or more thin wires, as it usu­ally is. In that case, a vol­ume el­e­ment of nonzero cur­rent dis­tri­b­u­tion can be writ­ten as

$$\vec\jmath { \rm d}^3 {\underline{\skew0\vec r}}= I { \rm d}{\underline{\skew0\vec r}}$$

where in the right hand side ${\underline{\skew0\vec r}}$ de­scribes the po­si­tion of the cen­ter­line of the wire and $I$ is the cur­rent through the wire. More specif­i­cally, $I$ is the in­te­gral of the cur­rent den­sity over the cross sec­tion of the wire. The Biot-Savart law be­comes

$$\skew2\vec{\cal B}= - \frac{1}{4\pi\epsilon_0c^2} \int \fr... ...\underline{\skew0\vec r}}) { \rm d}{\underline{\skew0\vec r}}$$

where the in­te­gra­tion is over all in­fin­i­tes­i­mal seg­ments ${ \rm d}{\underline{\skew0\vec r}}$ of the wires.