Question

A steady current $I$ flows through a wire of radius $r$, length $L$ and resistivity $\rho$. The current produces heat in the wire. The rate of he at loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of

• A. $L$
• B. $r$
• C. $I$
• D. $\rho$

Answer 1

Answer: (A)

Given, heat loss per second $Q_1$ through wire is proportional to surface area of wire.
$\Rightarrow Q_1=k\pi r^{2}L\left(\Delta T\right)$
where, $k=$ proportionality constant
and $\Delta T$ temperature difference of wire and surroundings.
and heat generated per second is
$Q_2=I^{2}R=\frac{I^2\rho L}{\pi r^2}$
In steady state, $Q_1=Q_2$
$\Rightarrow k\pi r^{2}L\left(\Delta T\right)=\frac{I^2\rho L}{\mu r^2}$
$\Rightarrow \Delta T=\frac{I^2\rho }{k{\pi }^2{r}^4}$
$\therefore$ Steady state temperature is independent of length of wire