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(C) Given: Length $L = 10 \, m$,Radius $r = \frac{10^{-2}}{\sqrt{\pi}} \, m$,Resistance $R = 10 \, \Omega$,Electric field $E = 10 \, V/m$. First,calcu

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$10^{5} \, A/m^{2}$

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(C) Given: Length $L = 10 \, m$,Radius $r = \frac{10^{-2}}{\sqrt{\pi}} \, m$,Resistance $R = 10 \, \Omega$,Electric field $E = 10 \, V/m$. First,calculate the cross-sectional area $A$: $A = \pi r^{2} = \pi \left( \frac{10^{-2}}{\sqrt{\pi}} ight)^{2} = \pi \cdot \frac{10^{-4}}{\pi} = 10^{-4} \, m^{2}$. We know that current density $J = \sigma E$,where $\sigma$ is conductivity. Since $\sigma = \frac{1}{ho}$ and $R = ho \frac{L}{A}$,we have $ho = \frac{RA}{L}$. Thus,$\sigma = \frac{L}{RA}$. Substituting this into the expression for $J$: $J = \left( \frac{L}{RA} ight) E = \frac{E \cdot L}{R \cdot A}$. Plugging in the values: $J = \frac{10 	imes 10}{10 	imes 10^{-4}} = \frac{100}{10^{-3}} = 10^{5} \, A/m^{2}$.

$A$ copper wire of length $10 \, m$ and radius $(10^{-2} / \sqrt{\pi}) \, m$ has an electrical resistance of $10 \, \Omega$. The current density in the wire for an electric field strength of $10 \, V/m$ is:

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