A copper wire of length 10m and radius <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mo>/</mo> <mi>π</mi> </mrow> </mrow> </mfenced> <mi mathvariant="normal">m</mi> </math> has electrical resistance of <math> <mn>10</mn> <mi mathvariant="normal">Ω</mi> </math> . The current density in the wire for an electric field strength of <math> <mn>10</mn> <mo>(</mo> <mtext> </mtext> <mi mathvariant="normal">V</mi> <mo>/</mo> <mi mathvariant="normal">m</mi> <mo>)</mo> </math> is:

J = σ E ⇒ E ρ = E l R A = 10 × 10 × π 10 × 10 - 4 ⇒ 10 5 A / m 2

A copper wire of length 10m and radius <math> <mfenced separators="|"> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>2</mn> </mrow> </mrow> </msup> <mo>/</mo> <mi>π</mi> </mrow> </mrow> </mfenced> <mi mathvariant="normal">m</mi> </math> has electrical resistance of <math> <mn>10</mn> <mi mathvariant="normal">Ω</mi> </math> . The current density in the wire for an electric field strength of <math> <mn>10</mn> <mo>(</mo> <mtext> </mtext> <mi mathvariant="normal">V</mi> <mo>/</mo> <mi mathvariant="normal">m</mi> <mo>)</mo> </math> &nbsp;is: &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

Ask & Answer Question

Option 1 -

$10^{4} A / m^{2}$

Option 2 -

$10^{6} A / m^{2}$

Option 3 -

$10^{- 5} A / m^{2}$

Option 4 -

$10^{5} A / m^{2}$

2 Views | Posted a month ago

Answered by

Raj Pandey | Contributor-Level 9

a month ago

Correct Option - 4

Detailed Solution:

$J = σ E \Rightarrow \frac{E}{ρ} = \frac{E l}{R A} = \frac{10 \times 10 \times π}{10 \times 10^{- 4}}$

$\Rightarrow 10^{5} A / m^{2}$

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