A steady current I flows through a wire of ra | vedclass

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Question

(a)

Given, heat loss per second $Q_1$ through wire is proportional to surface area of wire.

$\Rightarrow \quad Q_1=k \pi r^2 L(\Delta T)$

whe

Vedclass · Accepted Answer

$L$

Vedclass · Answer

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

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 heat loss in a wire is proportional to its surface area. The steady temperature of the wire is independent of

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