Question

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

Answer 1

Using Ampere's Circuital Law,

${\oint }_{p}B.dl={\mu }_{o}I$ (Equation 1)

Total current ($I$) flowing through the cross-section of radius $r$, $I={\int }_0^{r}j\left(r^{\prime }\right)\times A$

$I={\int }_0^r\frac{xb(\alpha +1)(r^{\prime }{)}^{\alpha -1}}{{\mu }_o}2\pi r^{\prime }d{r}^{\prime }$

$I=\frac{xb(\alpha +1)}{{\mu }_o}2\pi {\int }_0^r(r^{\prime }{)}^{\alpha }d{r}^{\prime }$

$I=\frac{xb(\alpha +1)}{{\mu }_o}2\pi \frac{r^{\alpha +1}}{\alpha +1}$

$I=\frac{bx2\pi r^{\alpha +1}}{{\mu }_o}$

${\oint }_{p}B.dl={\oint }_{p}b{r}^{\alpha }dl=b{r}^{\alpha }{\oint }_{p}dl$

${\oint }_{p}B.dl=b{r}^{\alpha }2\pi r=2\pi b{r}^{\alpha +1}$

From equation 1 , $2\pi b{r}^{\alpha +1}={\mu }_o\frac{bx2\pi r^{\alpha +1}}{{\mu }_o}$

Hence, $x=1$