Step-by-step Derivation

1. Determine the Electric Field inside the Cylinder: Let's consider the system at an arbitrary time t > 0. The charge on the infinite line is no longer constant because it leaks away through the conducting material. Let the linear charge density at time t be λ(t). To find the electric field E at a radial distance r from the axis (for 0 < r < R), we use Gauss's Law. Consider a cylindrical Gaussian surface of radius r and length L coaxial with the line charge. $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon}$ The electric field is radial due to symmetry. So, $E(r, t) \cdot (2\pi r L) = \frac{\lambda(t) L}{\epsilon}$. This gives the electric field at any point inside the material: $E(r, t) = \frac{\lambda(t)}{2\pi\epsilon r}$ The direction of the field is radially outward.

2. Apply Ohm's Law to find Current Density: The problem states that the material follows Ohm's law, $\vec{j} = \sigma \vec{E}$, where $\sigma$ is the electrical conductivity. The current density j(r, t) is therefore also radial and its magnitude is: $j(r, t) = \sigma E(r, t) = \frac{\sigma \lambda(t)}{2\pi\epsilon r}$ This equation shows that at any given time t, the current density varies with r as 1/r. However, the question asks for the time variation of j at any point. The time dependence of j at any r is determined by the time dependence of λ(t).

3. Relate Current to the Rate of Change of Charge: The current flowing radially outward through a cylindrical surface of radius r and length L is given by $I(t) = j(r, t) \times Area$. $I(t) = \left( \frac{\sigma \lambda(t)}{2\pi\epsilon r} \right) \times (2\pi r L) = \frac{\sigma \lambda(t) L}{\epsilon}$ This current I(t) is due to the charge leaving the central line charge. If Q(t) = λ(t)L is the charge on a length L of the line, then the current is the rate of decrease of this charge: $I(t) = -\frac{dQ(t)}{dt} = -\frac{d(\lambda(t)L)}{dt} = -L \frac{d\lambda(t)}{dt}$

4. Form and Solve the Differential Equation for λ(t): Equating the two expressions for I(t): $\frac{\sigma \lambda(t) L}{\epsilon} = -L \frac{d\lambda(t)}{dt}$ $\frac{d\lambda(t)}{dt} = -\left(\frac{\sigma}{\epsilon}\right) \lambda(t)$ This is a first-order linear differential equation, which describes exponential decay. The solution is: $\lambda(t) = \lambda_0 e^{-(\sigma/\epsilon)t}$ where $\lambda_0$ is the initial linear charge density at t = 0.

5. Determine the Time Variation of Current Density j(t): Now, we substitute the expression for λ(t) back into the equation for j(r, t): $j(r, t) = \frac{\sigma}{2\pi\epsilon r} \lambda(t) = \frac{\sigma}{2\pi\epsilon r} (\lambda_0 e^{-(\sigma/\epsilon)t})$ We can write this as: $j(r, t) = j(r, 0) e^{-t/\tau}$ where $j(r, 0) = \frac{\sigma \lambda_0}{2\pi\epsilon r}$ is the initial current density at radius r, and $\tau = \epsilon/\sigma$ is the relaxation time constant of the material.

6. Analyze the Result and Choose the Correct Graph: The equation $j(r, t) = j(r, 0) e^{-t/\tau}$ shows that the magnitude of the current density at any point r decreases exponentially with time.

   • At t = 0, the current density is maximum, j = j(r, 0).
   • As $t \to \infty$, the current density $j \to 0$. Let's examine the given graphs:
   • Graph A shows a constant current density. Incorrect.
   • Graph B shows a linearly decreasing current density. Incorrect.
   • Graph C shows the current density starting from zero, which is incorrect. The current is maximum at t=0 when the charge and electric field are maximum.
   • Graph D shows a quantity that starts at a maximum value at t=0 and decays exponentially towards zero. This perfectly matches our derived function for j(t).

Therefore, graph D best describes the variation of the magnitude of current density with time.