A current I flows through a cylindrical rod of uniform cross-sectional area A and resistivity ρ. The electric flux through the shaded cross-section of rod as shown in figure is:

Question

A current I flows through a cylindrical rod of uniform cross-sectional area A and resistivity ρ. The electric flux through the shaded cross-section of rod as shown in figure is: (A) (I/ρ) (B) ρ I (C) (ρ I/A) (D) (ρ A/I)

Tardigrade Admin · Accepted Answer

The current density J, electric field E at the cross section are related by J=(E/ρ) Multiplying both side by A, we get J A=(E A/ρ) or I=(φ/ρ) where φ is electric flux. or φ=ρ I

Q. A current $I$ flows through a cylindrical rod of uniform cross-sectional area $A$ and resistivity $ρ$ . The electric flux through the shaded cross-section of rod as shown in figure is:

$\frac{I}{ρ}$

$ρ I$

$\frac{ρ I}{A}$

$\frac{ρ A}{I}$

The current density $J$ , electric field $E$ at the cross section are related by $J = \frac{E}{ρ}$
Multiplying both side by $A$ , we get
$J A = \frac{E A}{ρ}$ or $I = \frac{ϕ}{ρ}$ where $ϕ$ is electric flux.
or $ϕ = ρ I$

Q. A current I flows through a cylindrical rod of uniform cross-sectional area A and resistivity ρ. The electric flux through the shaded cross-section of rod as shown in figure is:

ρI​

ρI

AρI​

IρA​