QJEEAdv16Phc10a: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phc10a Solution: An infinite line charge of uniform electric charge density l…

Let the charge per unit length on the axis be $\lambda$ (t) at a time t. The electric field due to this line charge at a radial distance r is given by

$\overrightarrow E (r,t) = {{\lambda (t)} \over {2\pi \in r}}\widehat r$ . ...... (1)

The free charges inside the conductor start moving radially due to the presence of electric field. The current density at a point is defined as the current flowing across a unit area placed perpendicular to the direction of current flow. The current density at a point is related to the electric field at that point by Ohm's law i.e.,

$\overrightarrow j (r,t) = \sigma \overrightarrow E (r,t)$ , ....... (2)

where $\sigma$ is the electrical conductivity of the conductor. The current through a cylindrical shell of radius r and length l is given by

$I = \int_{surface} {\overrightarrow j (r,t)\,.\,d\overrightarrow A = j(r,t)(2\pi rl)}$ . ...... (3)

By conservation of charge, the current I is equal to the rate of decrease of charge q on axial line segment of length l i.e.,

$I = - {{dq} \over {dt}} = - l{{d\lambda (t)} \over {dt}}$ . ...... (4)

From equations (1) - (4)

${{d\lambda (t)} \over {dt}} = - {{\sigma \lambda (t)} \over \in }$ .

Integrate with initial condition $\lambda$ (t = 0) = $\lambda$ ₀ to get

$\lambda (t) = {\lambda _0}\exp {e^{ - {{\sigma t} \over \in }}}$ ,

Substitute $\lambda$ (t) in equation (2) to get

$\overrightarrow j (r,t) = {{\sigma {\lambda _0}} \over {2\pi \in r}}{e^{ - {{\sigma t} \over \in }}}$ .

Note that current density varies as 1/r with radial distance r. It decreases exponentially with time and becomes zero at t $\to$ $\infty$ .