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 $j(r) = \dfrac{{xb(\alpha + 1)}}{{{\mu _0}}}{r^{(\alpha - 1)}}$ if the magnetic induction inside the stream varies as $B = b{r^\alpha }$, where $b$ and $\alpha$ are positive constants. Find $x$

Answer 1

Hint
This question is based on the concept of the current density of the parallel stream of electrons and related magnetic induction. We can use the ampere's circuital law and on equating both the sides we will get the value of $x$
In this solution we will be using the following formula,
$\Rightarrow \oint {\vec B \cdot d\vec l} = {\mu _o}I$
where $\vec B$ is the magnetic field, ${\mu _o}$ is the permittivity and $I$ is the current

Complete step by step answer
In the question we are given the current density. Let us consider the electrons are flowing in a stream of radius $r$. We consider a thin ring on the surface area of the stream having radius $r'$. So the area of this ring is,
$\Rightarrow A = 2\pi r'dr'$
So the current in this ring is the product of the current density and the area,
$\therefore dI = j\left( {r'} \right)2\pi r'dr'$
So for the whole current we integrate in the range from 0 to $r$. Hence we get,
$\Rightarrow I = \int {dI}$
Substituting the values and the limit we have,
$\Rightarrow I = \int\limits_0^r {j\left( {r'} \right)} 2\pi r'dr'$
In the question we are given
$\Rightarrow j(r) = \dfrac{{xb(\alpha + 1)}}{{{\mu _0}}}{r^{(\alpha - 1)}}$
Substituting in the equation we get
$\Rightarrow I = \int\limits_0^r {\dfrac{{xb(\alpha + 1)}}{{{\mu _0}}}{{r'}^{(\alpha - 1)}}2\pi r'dr'}$
Taking all the constants out of the integration
$\Rightarrow I = \dfrac{{xb(\alpha + 1)2\pi }}{{{\mu _0}}}\int\limits_0^r {{{r'}^{(\alpha - 1)}}r'dr'}$
Therefore we have,
$\Rightarrow I = \dfrac{{xb(\alpha + 1)2\pi }}{{{\mu _0}}}\int\limits_0^r {{{r'}^\alpha }dr'}$
On integrating we get,
$\Rightarrow I = \dfrac{{xb(\alpha + 1)2\pi }}{{{\mu _0}}}\left. {\dfrac{{{{r'}^{\left( {\alpha + 1} \right)}}}}{{\left( {\alpha + 1} \right)}}} \right|_0^r$
On substituting the limits,
$\Rightarrow I = \dfrac{{xb(\alpha + 1)2\pi }}{{{\mu _0}\left( {\alpha + 1} \right)}}\left[ {{r^{\left( {\alpha + 1} \right)}}} \right]$
On cancelling $\left( {\alpha + 1} \right)$ from numerator and denominator,
$\Rightarrow I = \dfrac{{xb2\pi }}{{{\mu _0}}}\left[ {{r^{\left( {\alpha + 1} \right)}}} \right]$
Now we are given $B = b{r^\alpha }$
So the integration $\oint {\vec B \cdot d\vec l}$ gives,
$\Rightarrow \oint {\vec B \cdot d\vec l} = \oint {b{r^\alpha }d\vec l}$
Taking the constants out and integrating we get,
$\Rightarrow \oint {\vec B \cdot d\vec l} = b{r^\alpha }\oint {d\vec l}$
The line integration gives,
$\Rightarrow \oint {\vec B \cdot d\vec l} = b{r^\alpha }2\pi r$
From the ampere’s circuital law,
$\Rightarrow \oint {\vec B \cdot d\vec l} = {\mu _o}I$
So substituting the values we have,
$\Rightarrow b{r^\alpha }2\pi r = {\mu _0}\dfrac{{xb2\pi }}{{{\mu _0}}}\left[ {{r^{\left( {\alpha + 1} \right)}}} \right]$
Cancelling the same terms from both sides,
$\Rightarrow {r^\alpha }r = x\left[ {{r^{\left( {\alpha + 1} \right)}}} \right]$
Therefore, we can write from the above equation $x = 1$.

Note
The Ampere’s circuital law in electromagnetism relates the integrated magnetic field in a closed loop to the current passing through that loop with a constant of proportionality called the permittivity in free space. It is given by the form, $\oint {\vec B \cdot d\vec l} = {\mu _o}I$.