The current density in conducting wire is $8.4 A / m^{2}$ .Potential gradient along wire is 7 V/cm then conductivity of material (in SI units)

Ω^{- 1} m

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Ω^{- 1} m

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120

Ω^{- 1} m

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Ω^{- 1} m

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Solution

The correct option is C 120 $Ω^{- 1} m$
7 V/cm= $\frac{7}{10^{- 2}} = 700$ V/m.
Now, current density is given by
$j = σ E$
$\Rightarrow σ = \frac{j}{E} = \frac{8.4 \times 10^{4} a m p / m^{2}}{700 V / m}$
$= \frac{8.4 \times 100 \times 10 \times 10}{700} = 120$ $Ω^{- 1} m$

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The current density in conducting wire is 8.4 A/m2 .Potential gradient along wire is 7 V/cm then conductivity of material (in SI units)

The correct option is C 120 Ω−1m7 V/cm= 710−2=700 V/m. Now, current density is given by j=σE ⇒σ=jE=8.4×104amp/m2700V/m =8.4×100×10×10700=120 Ω−1m

The current density in conducting wire is $8.4 A / m^{2}$ .Potential gradient along wire is 7 V/cm then conductivity of material (in SI units)

The correct option is C 120 $Ω^{- 1} m$
7 V/cm= $\frac{7}{10^{- 2}} = 700$ V/m.
Now, current density is given by
$j = σ E$
$\Rightarrow σ = \frac{j}{E} = \frac{8.4 \times 10^{4} a m p / m^{2}}{700 V / m}$
$= \frac{8.4 \times 100 \times 10 \times 10}{700} = 120$ $Ω^{- 1} m$