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(A) The rate of heat generated in the wire is given by $Q_2 = I^2 R$. Since resistance $R = \frac{\rho L}{\pi r^2}$,we have $Q_2 = \frac{I^2 \rho L}{\

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(A) The rate of heat generated in the wire is given by $Q_2 = I^2 R$. Since resistance $R = \frac{\rho L}{\pi r^2}$,we have $Q_2 = \frac{I^2 \rho L}{\pi r^2}$.The rate of heat loss $Q_1$ is proportional to the surface area $A = 2 \pi r L$. Thus,$Q_1 = k (2 \pi r L) \Delta T$,where $k$ is a constant and $\Delta T$ is the temperature difference.In a steady state,the rate of heat generated equals the rate of heat loss: $Q_2 = Q_1$.$\frac{I^2 \rho L}{\pi r^2} = k (2 \pi r L) \Delta T$.Solving for $\Delta T$: $\Delta T = \frac{I^2 \rho L}{\pi r^2 \cdot 2 \pi r L k} = \frac{I^2 \rho}{2 \pi^2 r^3 k}$.Since the length $L$ cancels out from both sides,the steady-state temperature difference $\Delta T$ is independent of the length $L$ of the 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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