Q6P - Fundamentals Of Physics

Question

A certain cylindrical wire carries current. We draw a circle of radius r around its central axis in figure-a to determine the current i within the circle. Figure-b shows current i as a function of r². The vertical scale is set by ${\mathit{i}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{m}\mathit{A}$ ,and the horizontal scale is set by, ${\mathit{r}}_{\mathbf{s}}^{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{m}{\mathit{m}}^{\mathbf{2}}$ . (a) Is the current density uniform?  (b) If so, what is its magnitude?

 

                         

Step-by-Step Solution

Answer

1. The current density is uniform.
2. The magnitude of the current density is $4.0\times {10}^2 A/m^2$' .

Step 1: The given data

From graph, 

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

Step 2: Understanding the concept of the current density

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, $J=\frac{i}{A}$                            …(i)

The area of the circle, $A={\mathrm{\pi }}^2$                                                                                    …(ii)

Step 3: (a) Calculation of the behavior of the current density

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 $\mathrm{\pi }$. From the graph, we can see that current increases linearly, that means,

  $$\begin{gathered} \frac{i}{A}=\frac{i}{{\mathrm{\pi r}}^2} \\ =cons\tan t \end{gathered}$$

Hence, we can say that current density is uniform.

Step 4: (b) Calculation of the magnitude of the current density

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

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

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