Answer to Question #135621 in Physics for Sridhar

Question

A circular wire has current density
J=[2×10^10(A)/(m^2)]r^2 where r is the radial distance out of the wire radius is 2mm .The end-to-end potential applied to the wire is 50V . How much energy (in joule) is converted to thermal energy in 100s ?
Ans 800πJ

Answer 1

By definition, the total current in the wire can be found as following:

$$I = \int_0^r Jdr$$

where $r = 2mm = 0.002m$ is the radius of the wire. Obtain:

$$I = 2\times 10^{10}\int_0^{0.002} r^2dr = 2\times 10^{10}\dfrac{r^3}{3}|^{0.002}_0 = \dfrac{160}{3}A$$

The energy converted to the Joule heating is:

$$Q = VIt$$

where $V = 50V$, $t = 100s$. Thus, get:

$$Q = VIt = 50\cdot \dfrac{160}{3}\cdot 100 \approx 266666.67J\approx 0.27MJ$$

Answer. 266666.67 J or 0.27 MJ.