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A cylindrical resistor of radius $7.0 \mathrm{~mm}$ and length 4.0 $\mathrm{cm}$ is made of material that has a resistivity of $10^{-6} \Omega . \mathrm{m}$. If the energy is dissipated at rate $1.54 \mathrm{~W}$ in the resistor, then the current density is
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The correct answer is:
$5 \times 10^5 \mathrm{~A} / \mathrm{m}^2$
$P=i^2 R \Rightarrow i=\sqrt{\frac{P}{R}}=\sqrt{\frac{P A}{\rho l}}$
$\begin{aligned} & \text { So, } J=\frac{I}{A}=\sqrt{\frac{P}{\rho l A}} \\ & =\sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times\left(7 \times 10^{-3}\right)^2}} \\ & =5 \times 10^5 \mathrm{~A} / \mathrm{m}^2\end{aligned}$
$\begin{aligned} & \text { So, } J=\frac{I}{A}=\sqrt{\frac{P}{\rho l A}} \\ & =\sqrt{\frac{1.54}{10^{-6} \times 0.04 \times \pi \times\left(7 \times 10^{-3}\right)^2}} \\ & =5 \times 10^5 \mathrm{~A} / \mathrm{m}^2\end{aligned}$
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