91Ó°ÊÓ

a) Current density,$J_0=5.50×{10}^4\mathrm{A}/{\mathrm{m}}^2$

b) Radius of the wire,$\mathrm{R}=3.40\mathrm{mm}\mathrm{or}3.40×{10}^{-3}\mathrm{m}$

The term "current density" refers to the quantity of electric current moving across a certain cross-section. We use the relation between the current and the current density to find the current in the wire. After finding the current for different current densities, we can check which function maximizes the current density.

Formulae:

The equation of the current flowing through a small area,$i=∫\stackrel{→}{J}.\stackrel{→}{dA}$ ...(i)

The cross-sectional area of the circle, $A=\mathrm{Ï€}{r}^2$ ...(ii)

We have, the given value of the current density as:$J_a=J_{0}r/R\mathrm{for}{J}_0=5.50×{10}^4$

The differential cross-sectional area value using equation (ii) can be given as follows:

$dA=2\mathrm{Ï€}rdr$

Substituting these above values in the equation (i), we can get the contained current within the width of the concentric ring assuming $\stackrel{→}{J}$is directed along the wire with varying radial distances from to $r=0tor=R$as follows:

$$\begin{gathered} i=∫\frac{J_{0}r}{R}2\mathrm{π}rdr \\ =\frac{2\mathrm{π}{J}_0}{R}{∫}_{r=0}^{r=R}{r}^{2}dr \\ =\frac{2\mathrm{π}{J}_0}{R}{\left[\frac{r^3}{3}\right]}_0^R \\ =\frac{2\mathrm{π}{J}_0}{R}\frac{R^3}{3} \\ =\frac{2\mathrm{π}{J}_0{R}^2}{3} \\ =\frac{2\mathrm{π}\left(5.5×{10}^{4}A/m^3\right){\left(3.40×{10}^{-3}m\right)}^2}{3} \\ =\frac{3.9948}{3}A \\ =1.33A \end{gathered}$$

Hence, the value of the current of this case is 1.33 A.

We have, the given value of the current density as: $J_b=J_0\left(1-\frac{r}{R}\right)\mathrm{for}{J}_0=5.50×{10}^4$for

The differential cross-sectional area value using equation (ii) can be given as follows:

$dA=2\mathrm{Ï€}rdr$

Substituting these above values in the equation (i), we can get the contained current within the width of the concentric ring assuming $\stackrel{→}{J}$is directed along the wire with varying radial distances from r = to r = R as follows:

$$\begin{gathered} i=∫J_0\left(1-\frac{r}{R}\right)2\mathrm{π}rdr \\ =2\mathrm{π}{J}_0\left[∫rdr-\frac{1}{R}∫r^{2dr}\right] \\ =2\mathrm{π}{J}_0{\left[\frac{r^2}{2}-\frac{1}{R}\frac{r^3}{3}\right]}_0^R \\ =2\mathrm{π}{J}_0\left[\frac{R^2}{2}-\frac{1}{R}\frac{R^3}{3}\right] \\ =\frac{2\mathrm{π}{J}_0{R}^2}{6} \\ =\frac{2\mathrm{π}\left(5.5×{10}^4\mathrm{A}/{\mathrm{m}}^2\right){\left(3.40×{10}^{-3}\mathrm{m}\right)}^3}{6} \\ =\frac{3.9948}{6} \\ =0.666\mathrm{A} \end{gathered}$$

Therefore, the value of the current flow is 0.666 A.

Current through the wire when the current density is $J_b$is different from that in part (a) because $J_b$is higher near the center of the cylinder, where the area is smaller for the same radial interval and it is lower outward, resulting in a lower average current density over the cross section and consequently, a lower current than that in part (a). Hence,$j_a$ has its maximum value near the surface of the wire.