Question

A long conductor of a circular cross-section of radius R is made of a material whose resistivity depends only on the distance 'r' from the conductor's central axis as $\rho = a/r^2$, where a is a constant. The conductor has length L (>>R). Its ends are maintained at a potential difference = $\Delta V$. Find (a) the electric field and current density in the conductor at any distance 'r' from the central axis.

(b) total rate of joule-heat production.

The rate of loss of heat from the surface of the conductor is $\lambda(T_s - T_A)$, where '$T_s$' is the surface temperature of the conductor, '$T_A$' the ambient temperature and '$\lambda$' is some constant. Find (c) surface temperature ($T_s$) in steady state.

If the thermal conductivity of the conductor's material is 'k' everywhere find (d) the temperature in the conductor as a function of 'r'. Assuming steady state.

Answer 1

(a) $E = \frac{\Delta V}{L}$ (uniform), $J(r) = \frac{\Delta V r^2}{aL}$ (along the length), (b) $P_{total} = \frac{\pi (\Delta V)^2 R^4}{2aL}$, (c) $T_s = T_A + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$, (d) $T(r) = T_A + \frac{(\Delta V)^2}{16akL^2}(R^4 - r^4) + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$

Answer 2

The problem asks for several quantities related to a long conductor with radius R and length L, made of a material with resistivity $\rho = a/r^2$, where r is the distance from the central axis. A potential difference $\Delta V$ is applied across the ends.

(a) Electric field and current density: Since the conductor is long (L >> R) and the potential difference is applied across the ends, the electric field $\vec{E}$ is uniform along the length of the conductor (say, along the z-axis) and its magnitude is given by $E = \frac{\Delta V}{L}$. The electric field is independent of r. The current density $\vec{J}$ is related to the electric field by Ohm's law: $\vec{E} = \rho \vec{J}$. Since $\vec{E}$ is along the z-axis and $\rho$ is a scalar function of r, $\vec{J}$ will also be along the z-axis. The magnitude of the current density at a distance r from the central axis is $J(r) = \frac{E}{\rho(r)}$. Substituting the expressions for E and $\rho(r)$: $J(r) = \frac{\Delta V/L}{a/r^2} = \frac{\Delta V}{L} \frac{r^2}{a} = \frac{\Delta V r^2}{aL}$. The electric field is $\vec{E} = \frac{\Delta V}{L} \hat{z}$ (assuming the conductor is along the z-axis). The current density is $\vec{J}(r) = \frac{\Delta V r^2}{aL} \hat{z}$.

(b) Total rate of joule-heat production: The rate of Joule heating per unit volume at a distance r is $P_v(r) = \vec{J}(r) \cdot \vec{E} = J(r) E$. $P_v(r) = \left(\frac{\Delta V r^2}{aL}\right) \left(\frac{\Delta V}{L}\right) = \frac{(\Delta V)^2 r^2}{aL^2}$. To find the total rate of heat production, we integrate the power density over the volume of the conductor. The volume element in cylindrical coordinates is $dV = (2\pi r dr) L$. Total power $P_{total} = \int_0^R P_v(r) (2\pi r dr L) = \int_0^R \frac{(\Delta V)^2 r^2}{aL^2} (2\pi r dr L)$. $P_{total} = \frac{2\pi (\Delta V)^2 L}{aL^2} \int_0^R r^3 dr = \frac{2\pi (\Delta V)^2}{aL} \left[\frac{r^4}{4}\right]_0^R = \frac{2\pi (\Delta V)^2}{aL} \frac{R^4}{4} = \frac{\pi (\Delta V)^2 R^4}{2aL}$.

Alternatively, we can calculate the total current $I = \int_0^R J(r) (2\pi r dr) = \int_0^R \frac{\Delta V r^2}{aL} (2\pi r dr) = \frac{2\pi \Delta V}{aL} \int_0^R r^3 dr = \frac{2\pi \Delta V}{aL} \frac{R^4}{4} = \frac{\pi \Delta V R^4}{2aL}$. The total power is $P_{total} = \Delta V \cdot I = \Delta V \left(\frac{\pi \Delta V R^4}{2aL}\right) = \frac{\pi (\Delta V)^2 R^4}{2aL}$.

(c) Surface temperature ($T_s$) in steady state: In steady state, the total rate of heat production equals the total rate of heat loss from the surface. The rate of loss of heat from the surface is given as $\lambda(T_s - T_A)$. Assuming this refers to the total rate of heat loss from the cylindrical surface (since L >> R, end effects are negligible). $P_{total} = \lambda(T_s - T_A)$. $\frac{\pi (\Delta V)^2 R^4}{2aL} = \lambda(T_s - T_A)$. $T_s - T_A = \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$. $T_s = T_A + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$.

(d) Temperature in the conductor as a function of 'r': In steady state, the heat generated within any volume must be conducted outwards. Consider a cylindrical shell of radius r and thickness dr and length L. The heat generated in this shell per unit length is $P_v(r) (2\pi r dr) = \frac{(\Delta V)^2 r^2}{aL^2} (2\pi r dr) = \frac{2\pi (\Delta V)^2 r^3 dr}{aL^2}$. The total heat generated per unit length within a cylinder of radius r is $Q'_{gen}(r) = \int_0^r \frac{2\pi (\Delta V)^2 r'^3 dr'}{aL^2} = \frac{2\pi (\Delta V)^2}{aL^2} \int_0^r r'^3 dr' = \frac{2\pi (\Delta V)^2}{aL^2} \frac{r^4}{4} = \frac{\pi (\Delta V)^2 r^4}{2aL^2}$. In steady state, this heat must be conducted outwards across the cylindrical surface at radius r (per unit length). According to Fourier's law of heat conduction, the radial heat flow rate per unit length across a cylindrical surface of radius r is $Q'_{cond}(r) = -k (2\pi r) \frac{dT}{dr}$. Equating the two rates: $Q'_{gen}(r) = Q'_{cond}(r)$. $\frac{\pi (\Delta V)^2 r^4}{2aL^2} = -k (2\pi r) \frac{dT}{dr}$. $\frac{dT}{dr} = - \frac{\pi (\Delta V)^2 r^4}{2aL^2} \frac{1}{2\pi k r} = - \frac{(\Delta V)^2 r^3}{4akL^2}$. Now, integrate with respect to r: $T(r) = \int - \frac{(\Delta V)^2 r^3}{4akL^2} dr = - \frac{(\Delta V)^2}{4akL^2} \int r^3 dr = - \frac{(\Delta V)^2}{4akL^2} \frac{r^4}{4} + C = - \frac{(\Delta V)^2 r^4}{16akL^2} + C$. We use the boundary condition that at the surface r=R, the temperature is $T_s$ from part (c). $T(R) = T_s = T_A + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$. $T(R) = - \frac{(\Delta V)^2 R^4}{16akL^2} + C$. So, $C = T_s + \frac{(\Delta V)^2 R^4}{16akL^2} = T_A + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda} + \frac{(\Delta V)^2 R^4}{16akL^2}$. Substituting C back into the expression for T(r): $T(r) = - \frac{(\Delta V)^2 r^4}{16akL^2} + T_A + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda} + \frac{(\Delta V)^2 R^4}{16akL^2}$. $T(r) = T_A + \frac{(\Delta V)^2 (R^4 - r^4)}{16akL^2} + \frac{\pi (\Delta V)^2 R^4}{2aL\lambda}$.