A cylindrical resistor of radius 7.0 , mm and | vedclass

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(B) The power dissipated in a resistor is given by $P = I^2 R$, where $I$ is the current and $R$ is the resistance.Resistance $R$ is given by $R = \rh

Vedclass · Accepted Answer

$5 	imes 10^5 \, A/m^2$

Vedclass · Answer

(B) The power dissipated in a resistor is given by $P = I^2 R$, where $I$ is the current and $R$ is the resistance.Resistance $R$ is given by $R = ho \frac{l}{A}$, where $ho = 10^{-6} \, \Omega \cdot m$, $l = 0.04 \, m$, and $A = \pi r^2 = \pi (7 	imes 10^{-3})^2 \, m^2$.Substituting $R$ in the power equation: $P = I^2 \left( \frac{ho l}{A} ight) \Rightarrow I = \sqrt{\frac{P A}{ho l}}$.Current density $J$ is defined as $J = \frac{I}{A}$.Substituting $I$: $J = \frac{1}{A} \sqrt{\frac{P A}{ho l}} = \sqrt{\frac{P}{ho l A}}$.Substituting the values: $J = \sqrt{\frac{1.54}{10^{-6} 	imes 0.04 	imes \pi 	imes (7 	imes 10^{-3})^2}}$.$J = \sqrt{\frac{1.54}{10^{-6} 	imes 0.04 	imes \pi 	imes 49 	imes 10^{-6}}} = \sqrt{\frac{1.54}{1.96 	imes 10^{-9} 	imes \pi}} \approx 5 	imes 10^5 \, A/m^2$.

$A$ cylindrical resistor of radius $7.0 \, mm$ and length $4.0 \, cm$ is made of a material that has a resistivity of $10^{-6} \, \Omega \cdot m$. If the energy is dissipated at a rate of $1.54 \, W$ in the resistor, then the current density is:

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