91Ó°ÊÓ

From graph,

1. The vertical scale is set by, i = 5.0 mA
2. The horizontal scale is set by,$r^2=4{\mathrm{mm}}^2$

The term "current density" refers to the quantity of electric current moving across a certain cross-section From the given graph, we can show that the current density is uniform. We can find the magnitude of the current density by substituting the given values in the formula of current density.

Formulae:

The current density of the current flowing through the area,$J=\frac{i}{A}$ …(¾±)

The area of the circle, $A={\mathrm{Ï€}}^2$ …(¾±¾±)

From equation (ii), we see that the area is directly proportional to $r^2$.

The horizontal axis of the graph is effectively same as the area enclosed at various radius values, except for a factor of$\mathrm{Ï€}$. From the graph, we can see that current increases linearly, that means,

role="math" localid="1661411499862" $$\begin{gathered} \frac{i}{A}=\frac{i}{{\mathrm{Ï€°ù}}^2} \\ =cons\tan t \end{gathered}$$

Hence, we can say that current density is uniform.

Substituting the value of area from equation (ii) in equation (i), we can get the magnitude value of the current density as follows:

$$\begin{gathered} J=\frac{5mA}{\mathrm{π}×4{\mathrm{mm}}^2} \\ =\frac{5×{10}^{-3}A}{\mathrm{π}×4×{10}^{-6}{\mathrm{m}}^2} \\ =397.88A/m^2 \\ ≈4.0×{10}^{2}A/m^2 \end{gathered}$$

Hence, the value of the current density is $4.0×{10}^{2}A/m^2$.