Question

Find the current density as a function of distance $r$ from the axis of a radially symmetrical parallel stream of electrons if the magnetic induction inside the stream varies as $B=b{r}^a$, where $b$ and $a$ are positive constants. if your answer is given by $j\left(r\right)=\frac{x\ast b(\alpha +1)}{{\mu }_0}{r}^{\alpha -1}$. Find x.

Answer 1

From Ampere's law: $B_{\varphi }2\pi r={\mu }_0{\int }_0^{r}j\left(r^{\prime }\right)2\pi r^{\prime }d{r}^{\prime }$
$j\left(r^{\prime }\right)=$ current density at a distance $r^{\prime }$
Given: $B_{\varphi }=b{r}^{\alpha }$
Substituting in above, $b{r}^{\alpha +1}={\mu }_0\int j\left(r^{\prime }\right)r^{\prime }d{r}^{\prime }$
Differentiating, we obtain: $(\alpha +1)b{r}^{\alpha }={\mu }_{0}j\left(r\right)r$
$\Rightarrow j\left(r\right)=\frac{b(\alpha +1)}{{\mu }_0}{r}^{\alpha -1}$