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