The field $\FLPH$
Next, we want to insert the current as written in Eq. (36.10) into Maxwell’s equations. We get  \begin{equation*} c^2\FLPcurl{\FLPB}=\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}= \frac{1}{\epsO}\biggl( \FLPj_{\text{cond}}+\FLPcurl{\FLPM}+\ddp{\FLPP}{t} \biggr)+\ddp{\FLPE}{t}. \end{equation*} 
 \begin{align*} c^2\FLPcurl{\FLPB}&=\frac{\FLPj}{\epsO}+\ddp{\FLPE}{t}\\[1ex] &=\frac{1}{\epsO}\biggl(\! \FLPj_{\text{cond}}\!+\!\FLPcurl{\FLPM}\!+\!\ddp{\FLPP}{t} \!\biggr)\!+\!\ddp{\FLPE}{t}. \end{align*}  We can move the term in $\FLPM$ to the left-hand side:  \begin{equation} \label{Eq:II:36:11} c^2\FLPcurl{\biggl(\FLPB-\frac{\FLPM}{\epsO c^2}\biggr)}=\, \frac{\FLPj_{\text{cond}}}{\epsO}+\ddp{}{t}\biggl( \FLPE+\frac{\FLPP}{\epsO}\biggr). \end{equation} 
 \begin{equation} \label{Eq:II:36:11} c^2\FLPcurl{\biggl(\!\FLPB\!-\!\frac{\FLPM}{\epsO c^2}\!\biggr)}\!= \frac{\FLPj_{\text{cond}}}{\epsO}\!+\!\ddp{}{t}\biggl(\! \FLPE\!+\!\frac{\FLPP}{\epsO}\!\biggr). \end{equation}  As we remarked in Chapter 32, many people like to write $(\FLPE+\FLPP/\epsO)$ as a new vector field $\FLPD/\epsO$. Similarly, it is often convenient to write $(\FLPB-\FLPM/\epsO c^2)$ as a single vector field. We choose to define a new vector field $\FLPH$ by \begin{equation} \label{Eq:II:36:12} \FLPH=\FLPB-\frac{\FLPM}{\epsO c^2}. \end{equation} Then Eq. (36.11) becomes \begin{equation} \label{Eq:II:36:13} \epsO c^2\FLPcurl{\FLPH}=\FLPj_{\text{cond}}+\ddp{\FLPD}{t}. \end{equation}  It looks simple, but all the complexity is just hidden in the letters $\FLPD$ and $\FLPH$. Now we have to give you a warning. Most people who use the mks units have chosen to use a different definition of $\FLPH$. Calling their field $\FLPH'$ (of course, they still call it $\FLPH$ without the prime), it is defined by \begin{equation} \label{Eq:II:36:14} \FLPH'=\epsO c^2\FLPB-\FLPM. \end{equation}  (Also, they usually write $\epsO c^2$ as a new number $1/\mu_0$; then they have one more constant to keep track of!) With this definition, Eq. (36.13) looks even simpler: \begin{equation} \label{Eq:II:36:15} \FLPcurl{\FLPH'}=\FLPj_{\text{cond}}+\ddp{\FLPD}{t}. \end{equation}  But the difficulties with this definition of $\FLPH'$ are, first, that it doesn’t agree with the definition of people who don’t use the mks units, and second, that it makes $\FLPH'$ and $\FLPB$ have different units. We think it is more convenient for $\FLPH$ to have the same units as $\FLPB$—rather than the units of $\FLPM$, as $\FLPH'$ does. But if you are going to be an engineer and work on the design of transformers, magnets, and such, you will have to watch out. You will find many books which use for $\FLPH$ the definition of Eq. (36.14) rather than our definition of Eq. (36.12), and many other books—especially handbooks about magnetic materials—that relate $\FLPB$ and $\FLPH$ the way we have done. You’ll have to be careful to figure out which convention they are using. One way to tell is by the units they use. Remember that in the mks system, $\FLPB$—and therefore our $\FLPH$—are measured with the unit: one weber per square meter, equal to $10{,}000$ gauss. In the mks system, a magnetic moment (a current times an area) has the unit: one ampere-meter$^2$. The magnetization $\FLPM$, then, has the unit: one ampere per meter. For $\FLPH'$ the units are the same as for $\FLPM$. You can see that this also agrees with Eq. (36.15), since $\FLPnabla$ has the dimensions of one over a length. People who are working with electromagnets also get in the habit of calling the unit of $\FLPH$ (with the $\FLPH'$ definition) “one ampere turn per meter”—thinking of the turns of wire on a winding. But a “turn” is really a dimensionless number, so that doesn’t need to confuse you. Since our $H$ is equal to $H'/\epsO c^2$, if you are using the mks system, $H$ (in webers/meter$^2$) is equal to $4\pi\times10^{-7}$ times $H'$ (in amperes per meter). It is perhaps more convenient to remember that $H$ (in gauss)${}=0.0126H'$ (in amp/meter). There is one more horrible thing. Many people who use our definition of $\FLPH$ have decided to call the units of $\FLPH$ and $\FLPB$ by different names!  Even though they have the same dimensions, they call the unit of $\FLPB$ one gauss, and the unit of $\FLPH$ one oersted (after Gauss and Oersted, of course). So, in many books you will find graphs with $\FLPB$ plotted in gauss and $\FLPH$ in oersteds. They are really the same unit—$10^{-4}$ of the mks unit. We have summarized the confusion about magnetic units in Table 36–1.