Variation of Current Density in Conducting Material Inside Cylindrical Shell

To determine the time-variation of the current density $j(t)$ within the conducting material, we analyze the system using Maxwell's equations and the equation of charge continuity.

1. Continuity Equation and Volume Charge Density

The electrical current density $\vec{j}$ in a medium with conductivity $\sigma$ obeys Ohm's law: $\vec{j} = \sigma \vec{E}$

By Gauss's law in a medium with permittivity $\varepsilon$, the volume charge density $\rho_v$ is related to the electric field $\vec{E}$ by: $\nabla \cdot \vec{E} = \frac{\rho_v}{\varepsilon}$

The continuity equation representing charge conservation is given by: $\nabla \cdot \vec{j} = -\frac{\partial \rho_v}{\partial t}$

Substituting $\vec{j} = \sigma \vec{E}$ into the continuity equation: $\sigma (\nabla \cdot \vec{E}) = -\frac{\partial \rho_v}{\partial t} \implies \frac{\sigma}{\varepsilon} \rho_v = -\frac{\partial \rho_v}{\partial t}$

Rearranging gives the differential equation for the volume charge density inside the bulk of the medium: $\frac{\partial \rho_v}{\partial t} + \frac{\sigma}{\varepsilon} \rho_v = 0$

Solving this differential equation yields: $\rho_v(r, t) = \rho_v(r, 0) e^{-\frac{\sigma}{\varepsilon} t}$

Since no initial volume charge density exists inside the bulk medium ($0 < r < R$) at $t = 0$, we have $\rho_v(r, 0) = 0$. Consequently, $\rho_v(r, t) = 0$ for all $0 < r < R$ at any time $t \ge 0$. Thus, charge flows radially outwards without accumulating inside the volume of the conductor.

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2. Decay of Line Charge Density

Since the charge density inside the volume remains zero, the electric field at a distance $r$ ($0 < r < R$) from the central axis depends entirely on the remaining line charge density $\lambda(t)$ at the central axis: $E(r, t) = \frac{\lambda(t)}{2\pi \varepsilon r}$

The total outward current per unit length flowing through a coaxial cylindrical surface of radius $r$ is: $I(t) = j(r, t) \cdot (2\pi r) = \sigma E(r, t) \cdot (2\pi r)$

By charge conservation on the line charge at $r = 0$, the rate of decrease of the line charge density equals the outward current per unit length: $I(t) = -\frac{d\lambda(t)}{dt}$

Substituting $E(r, t) = \frac{\lambda(t)}{2\pi \varepsilon r}$ into the expression for $I(t)$: $I(t) = 2\pi r \sigma \left(\frac{\lambda(t)}{2\pi \varepsilon r}\right) = \frac{\sigma}{\varepsilon} \lambda(t)$

Equating the two expressions for $I(t)$: $-\frac{d\lambda(t)}{dt} = \frac{\sigma}{\varepsilon} \lambda(t)$

Integrating with the initial condition $\lambda(0) = \lambda$: $\lambda(t) = \lambda e^{-\frac{\sigma}{\varepsilon} t}$

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3. Time Variation of Current Density

The current density $j(t)$ at any point at a distance $r$ from the central axis inside the material is given by Ohm's law: $j(t) = \sigma E(r, t) = \frac{\sigma \lambda(t)}{2\pi \varepsilon r}$

Substituting $\lambda(t) = \lambda e^{-\frac{\sigma}{\varepsilon} t}$: $j(t) = \left( \frac{\sigma \lambda}{2\pi \varepsilon r} \right) e^{-\frac{\sigma}{\varepsilon} t}$

Letting $j(0) = \frac{\sigma \lambda}{2\pi \varepsilon r}$ and defining the relaxation time $\tau = \frac{\varepsilon}{\sigma}$: $j(t) = j(0) e^{-t/\tau}$

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Conclusion

The magnitude of the current density $j(t)$ decreases exponentially over time, starting from a maximum initial value at $t = 0$ and asymptotically approaching zero as $t \to \infty$. This exponential decay curve is accurately depicted in Graph (C).

Correct Answer: C