2 A of current is flowing through a wire of resistance 5 - If the length of the wire is 4 m- then calculate the value of electric field inside the conductor-

Given : I=2 A R=5 Ω l=4 m We know that, J=σ E ⇒ E=Jσ E=IA×RAl [∵ J=IA and σ=1ρ=RAl] E=2×54=2.5 N/m ...

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Standard XII

Physics

Potential Gradient and Relation between Electric Field and Potential

2 A of curren...

Question

$2 A$ of current is flowing through a wire of resistance $5 Ω$ . If the length of the wire is $4 m$ , then calculate the value of electric field inside the conductor.

5 N/m

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1.5 N/m

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2.5 N/m

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8 N/m

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Solution

The correct option is C $2.5 N/m$
Given :
$I = 2 A$
$R = 5 Ω$
$l = 4 m$

We know that,
$J = σ E$

$\Rightarrow E = \frac{J}{σ}$

$E = \frac{I}{A} \times \frac{R A}{l}$
$[∵ J = \frac{I}{A} and σ = \frac{1}{ρ} = \frac{R A}{l}]$

$E = \frac{2 \times 5}{4} = 2.5 N/m$

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2 A of current is flowing through a wire of resistance 5 Ω. If the length of the wire is 4 m, then calculate the value of electric field inside the conductor.

The correct option is C 2.5 N/mGiven : I=2 A R=5 Ω l=4 m We know that, J=σE ⇒E=Jσ E=IA×RAl [∵J=IA and σ=1ρ=RAl] E=2×54=2.5 N/m

$2 A$ of current is flowing through a wire of resistance $5 Ω$ . If the length of the wire is $4 m$ , then calculate the value of electric field inside the conductor.

The correct option is C $2.5 N/m$
Given :
$I = 2 A$
$R = 5 Ω$
$l = 4 m$

We know that,
$J = σ E$

$\Rightarrow E = \frac{J}{σ}$

$E = \frac{I}{A} \times \frac{R A}{l}$
$[∵ J = \frac{I}{A} and σ = \frac{1}{ρ} = \frac{R A}{l}]$

$E = \frac{2 \times 5}{4} = 2.5 N/m$