A current of 5A is passing through a linear magnesium wire of cross-section 0.04 m2. The magnitude of electric field at every point of the conductor is (Take, resistivity of magnesium, ρ=44×10−8Ω−m)

Given, current, I = 5 AArea of cross-section of wire, A = 0.04 m2We know that, J=IA⇒ I=JAwhere, J = current density.⇒ 5=J4100 J=5004⇒ J=125 Am−2 The relation between electric field, current density and resistivity can be given asE=ρ⋅J=44×10−8×125 ∵ Resistivity, ρ=44×10−8Ω−m=5.5×10−5V/m

A current of 5A is passing through a linear magnesium wire of cross-section 0.04 m2. The magnitude of electric field at every point of the conductor is (Take, resistivity of magnesium, ρ=44×10−8Ω−m)

Given, current, I = 5 AArea of cross-section of wire, A = 0.04 m2We know that, J=IA⇒ I=JAwhere, J = current density.⇒ 5=J4100 J=5004⇒ J=125 Am−2 The relation between electric field, current density and resistivity can be given asE=ρ⋅J=44×10−8×125 ∵ Resistivity, ρ=44×10−8Ω−m=5.5×10−5V/m

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A current of 5A is passing through a linear magnesium wire of cross-section 0.04 m². The magnitude of electric field at every point of the conductor is (Take, resistivity of magnesium, $ρ = 44 \times 10^{- 8} Ω - m$ )

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$5.5 \times 10^{- 7} V / m$

$11 \times 10^{- 7} V / m$

$11 \times 10^{- 5} V / m$

$5.5 \times 10^{- 5} V / m$

answer is D.

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Detailed Solution

Given, current, I = 5 A

Area of cross-section of wire, A = 0.04 m²

We know that, $J = \frac{I}{A}$

$\Rightarrow I = JA$

where, J = current density.

$\begin{matrix} \Rightarrow 5 = J (\frac{4}{100}) \\ J = \frac{500}{4} \\ \Rightarrow J = 125 {Am}^{- 2} \end{matrix}$

The relation between electric field, current density and resistivity can be given as

$\begin{matrix} E = ρ \cdot J \\ = 44 \times 10^{- 8} \times 125 [∵ Resistivity, ρ = 44 \times 10^{- 8} Ω - m] \\ = 5.5 \times 10^{- 5} V / m \end{matrix}$