A wire of length Land cross-sectional area A has a resistivity that varies with distance x from one end as rho = rho_0(1 + alpha x/L). If a potential difference V is applied across the ends, what is the current density at a distance x?

B. V _0 L (1 + ) — The current I is constant throughout the wire. The resistance of a differential element dx is dR = rho dx / A. Integrating this gives total resistance R, and using I = V/R, the current density J = I/A is found to be independent of x.

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$A wire of length L and cross-sectional area A has a resistivity that varies with distance x from one end as ρ = ρ_{0} (1 + α x / L) . If a potential difference V is applied across the ends, what is the current density at a distance x ?$

A. $\frac{V}{ρ _{0} L ( 1 + α x / L )}$
B. $\frac{V α}{ρ _{0} L l n ( 1 + α )}$
C. $\frac{V}{ρ _{0} L}$
D. $\frac{V ( 1 + α x / L )}{ρ _{0} L}$

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Correct answer: B — $\frac{V α}{ρ _{0} L l n ( 1 + α )}$

Solution

The current I is constant throughout the wire. The resistance of a differential element d x is d R = \frac{ρ ( x ) d x}{A} . Integrating d R from 0 to L gives R = \frac{ρ _{0} L}{A} (1 + α /2) . However, since J = E / ρ and E = I / (σ A), we find J = \frac{V}{R A} = \frac{V α}{ρ _{0} L ln ( 1 + α )} is incorrect; the correct approach is J = \frac{V}{ρ ( x ) \int _{0}^{L} ρ ( x ) d x / A} .

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A wire of length L and cross-sectional area A has a resistivity that varies with distance x from one end as ρ=ρ0​(1+αx/L). If a potential difference V is applied across the ends, what is the current density at a distance x?

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$A wire of length L and cross-sectional area A has a resistivity that varies with distance x from one end as ρ = ρ_{0} (1 + α x / L) . If a potential difference V is applied across the ends, what is the current density at a distance x ?$