(A) The current density $J$ is related to the electric field $E$ by the relation $J = \frac{E}{\rho}$, where $\rho$ is the resistivity of the material

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(A) The current density $J$ is related to the electric field $E$ by the relation $J = \frac{E}{\rho}$, where $\rho$ is the resistivity of the material.We know that current density $J = \frac{I}{A}$.Therefore, $E = J \cdot \rho = \frac{I \cdot \rho}{A}$.Given values are $I = 5 \, A$, $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$, and $\rho = 3.0 \times 10^{-8} \, \Omega \cdot m$.Substituting these values into the formula:$E = \frac{5 \times 3.0 \times 10^{-8}}{1 \times 10^{-6}} = 15 \times 10^{-2} = 0.15 \, V/m$.

Class 12 Physics Current Electricity Current Density, Drift Velocity and Mobility

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$A$ conductor of length $100 \, cm$ and area of cross-section $1 \, mm^2$ carries a current of $5 \, A$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \, \Omega \cdot m$, then the electric field across the conductor is (in $ \, V/m$)

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A
$0.15$
B
$0.015$
C
$1.5$
D
$0.0015$

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Current Electricity

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Current Density, Drift Velocity and Mobility

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$A$ conductor of length $100 \, cm$ and area of cross-section $1 \, mm^2$ carries a current of $5 \, A$. If the resistivity of the material of the conductor is $3.0 \times 10^{-8} \, \Omega \cdot m$, then the electric field across the conductor is (in $ \, V/m$)

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