JEE Main · PhysicsHard

$A wire of length $L$ and 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 $x$?$

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Correct answer: B — $\frac{V \alpha}{\rho_0 L \ln(1 + \alpha)}$

Solution

The current

I

is constant throughout the wire. The resistance of a differential element

d x

d R = \frac{ρ ( x ) d x}{A}

. Integrating

d R

from

0

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}