A steady current $\mathit{I}$flows down a long cylindrical wire of radius a(Fig. 5.40). Find the magnetic field, both inside and outside the wire, if

1. The current is uniformly distributed over the outside surface of the wire.
2. The current is distributed in such a way that $\mathit{J}$is proportional to $\mathit{s}$,the distance from the axis.

a)

Calculate the magnetic field inside the cylindrical wire$(s<a)$

Write the expression for integral version of the Ampere’s Law.

$\mathbf{\oint }\mathit{B}\mathbf{\cdot }\mathit{d}\mathit{l}\mathbf{=}{\mathit{\mu }}_{\mathbf{0}}{\mathit{I}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$ …… (1)

Here,$I_{enc}$is the current enclosed by the Amperian loop and${\mu }_0$is the magnetic permeability in free vacuum.

Write the expression for enclosed current.

$I_{enc}=0$

Substitute $0$for $I_{enc}$, in equation (1)

$\begin{array}{l}B\oint dl={\mu }_0\left(0\right)\\ B\left(2\pi s\right)={\mu }_0\left(0\right)\\ B=0\end{array}$

Thus, the magnetic field inside the wire is 0.

Calculate the magnetic field outside the cylindrical wire$(s>a)$

Write the expression for enclosed current.

$I_{enc}=l$

Substitute$I$ for $I_{enc}$, in equation (1)

$\begin{array}{l}B\oint dl={\mu }_0\left(I\right)\\ B\left(2\pi s\right)={\mu }_0\left(I\right)\\ B=\frac{{\mu }_{0}I}{2\pi s}\end{array}$

Thus, the magnetic field outside the wire is$B=\frac{{\mu }_{0}I}{2\pi s}$ .

b)

Consider a point $s<a$,

Write the expression for current in terms of current density.

$I=\underset{0}{\overset{a}{\int }}J\cdot da$ …… (2)

Here,$J$is current density.

The current density is directly proportional to the distance from the axis $s$.

$\begin{array}{l}J\propto s\\ J=ks\end{array}$

Here, $k$is proportionality constant.

Substitute $ks$for $J$and $\left(2\pi s\right)ds$for $da$in equation (2)

Rearrange the above equation for $k$.

$K=\frac{3I}{2\pi a^3}$

Write the expression for the enclosed current in the region $s<a$.

$I_{enclosed}=\underset{0}{\overset{s}{\int }}J\cdot da$ …… (3)

Substitute$ks$for $J$and$\left(2\pi s\right)ds$for$da$in equation (3)

Now, substitute$\frac{3I}{2\pi a^3}$ for$K$in above equation.

Substitute$I\frac{s^3}{a^3}$ for $I_{enclosed}$in equation (1)

$\begin{array}{l}B\oint dl={\mu }_0\left(I\frac{s^3}{a^3}\right)\\ B\left(2\pi s\right)={\mu }_0\left(I\frac{s^3}{a^3}\right)\\ B=\frac{{\mu }_{0}I{s}^2}{2\pi a^3}\end{array}$

Thus, the magnetic field inside the wire is $B=\frac{{\mu }_{0}I{s}^2}{2\pi a^3}$.

Calculate the magnetic field outside the cylindrical wire$(s>a)$

Write the expression for enclosed current.

$I_{enc}=l$

Substitute$I$ for$I_{enc}$ , in equation (1)

$\begin{array}{l}B\oint dl={\mu }_0\left(I\right)\\ B\left(2\pi s\right)={\mu }_0\left(I\right)\\ B=\frac{{\mu }_{0}I}{2\pi s}\end{array}$

Thus, the magnetic field outside the wire is$B=\frac{{\mu }_{0}I}{2\pi s}$ .