91影视

a)

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

Write the expression for integral version of the Ampere鈥檚 Law.

$\mathbf{鈭�}\mathit{B}\mathbf{鈰�}\mathit{d}\mathit{l}\mathbf{=}{\mathit{渭}}_{\mathbf{0}}{\mathit{I}}_{\mathbf{e}\mathbf{n}\mathbf{c}}$ 鈥︹�� (1)

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

Write the expression for enclosed current.

$I_{enc}=0$

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

$\begin{array}{l}\text{鈥夆赌�}B鈭�dl={渭}_0\left(0\right)\\ B\left(2蟺s\right)={渭}_0\left(0\right)\\ \text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌�}B=0\end{array}$

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)

$\begin{array}{l}\text{鈥夆赌�}B鈭�dl={渭}_0\left(I\right)\\ B\left(2蟺s\right)={渭}_0\left(I\right)\\ \text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌�}B=\frac{{渭}_{0}I}{2蟺s}\end{array}$

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

b)

Consider a point $s<a$,

Write the expression for current in terms of current density.

$I=\underset{0}{\overset{a}{鈭�}}J鈰�da$ 鈥︹�� (2)

Here,$J$is current density.

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

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

Here, $k$is proportionality constant.

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

Rearrange the above equation for $k$.

$K=\frac{3I}{2蟺a^3}$

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

$I_{enclosed}=\underset{0}{\overset{s}{鈭�}}J鈰�da$ 鈥︹�� (3)

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

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

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

$\begin{array}{l}\text{鈥夆赌�}B鈭�dl={渭}_0\left(I\frac{s^3}{a^3}\right)\\ B\left(2蟺s\right)={渭}_0\left(I\frac{s^3}{a^3}\right)\\ \text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌�}B=\frac{{渭}_{0}I{s}^2}{2蟺a^3}\end{array}$

Thus, the magnetic field inside the wire is $\text{鈥�}B=\frac{{渭}_{0}I{s}^2}{2蟺a^3}$.

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)

$\begin{array}{l}\text{鈥夆赌�}B鈭�dl={渭}_0\left(I\right)\\ B\left(2蟺s\right)={渭}_0\left(I\right)\\ \text{鈥夆赌夆�夆赌夆�夆赌夆�夆赌�}B=\frac{{渭}_{0}I}{2蟺s}\end{array}$

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