thick slab extending from $z=-a$to $z=+a$(and infinite in the x andy directions) carries a uniform volume current $J=J\hat{x}$(Fig. 5.41). Find the magnetic field, as a function of $z$, both inside and outside the slab.

Consider the length and redraw the diagram of the slab.

Write the expression for Amperes law.

$\oint B·dI={\mu }_0{I}_{enc}$ …… (1)

Here, is the magnetic field, is the permeability in the vacuum, is the small element of length and is the enclosed by amperian loop.

Write the expression for the enclosed current in the region $0<z<a$.

$\begin{array}{rcl}{I}_{enc}& =& \oint J·da\\ & =& J\oint da\\ & =& JA\\ & & \end{array}$ …… (2)

Here, is the area of the Amperian loop.

Write the expression for the Amperian loop in the region $0<z<a$.

$A=Lz$

Substitute $Lz$for A in equation (2)

$I_{enc}=JzL$

Similarly,

Write the expression the enclosed current in the region $z>a$.

$\begin{array}{rcl}{I}_{enc}& =& \oint J·da\\ & =& JaL\\ & & \end{array}$

Use the Ampere’s law,

Write the expression for the magnetic field in the region $0<z<a$.

$$\begin{gathered} \oint B·dl={\mu }_0{I}_{enc} \\ BL={\mu }_0{I}_{enc} \\ B=\frac{{\mu }_0{I}_{enc}}{L} \end{gathered}$$

Substitute$JzL$for$I_{enc}$,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\left(JzL\right)}{L}\\ & =& {\mu }_{0}Jz\\ & & \end{array}$

Write the expression for line integral of magnetic field in the region $z>a$.

$\begin{array}{rcl}\oint B·dl& =& {\mu }_0{I}_{enc}\\ BL& =& {\mu }_0{I}_{enc}\\ B& =& \frac{{\mu }_0{I}_{enc}}{L}\\ & & \end{array}$

Substitute for $I_{enc}$,

$\begin{array}{rcl}B& =& \frac{{\mu }_0\left(JaL\right)}{L}\\ & =& {\mu }_{0}aJ\\ & & \end{array}$

According to right hand thumb rule, , magnetic field is directed towards negative y-axis.

Write the expression magnetic field $z>+a$.

$\begin{array}{rcl}B& =& {\mu }_{0}Ja\left(-\hat{y}\right)\\ & =& -{\mu }_{0}Ja\hat{y}\\ & & \end{array}$

Write the expression magnetic field $-a<z<a$.

$B=-{\mu }_{0}Jz\hat{y}$

Write the expression magnetic field$z>-a$

$\begin{array}{rcl}B& =& {\mu }_{0}J\left(-a\right)\left(-\hat{y}\right)\\ & =& {\mu }_{0}Ja\hat{y}\\ & & \end{array}$

Thus, the magnetic field inside the slab is $B=-{\mu }_{0}Jz\hat{y}$.

Thus, the magnetic field outside the slab for$z>+a$ is$B=-{\mu }_{0}Jz\hat{y}$ .

Thus, the magnetic field outside the slab for$z>-a$ is $B={\mu }_{0}Jz\hat{y}$.