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PhysicsCurrent ElectricityMicroscopicMedium2 minPYQ_2022

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A copper wire of length $10 m$ and radius $(\frac{10^{- 2}}{\sqrt{π}}) m$ has electrical resistance of $10 Ω$ . The current density in the wire for an electric field strength of $10 V m^{- 1}$ is:NEET 2022 (Phase 1)

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Answer:

Solution:

Length $= 10 m$

Radius $= (\frac{10^{- 2}}{\sqrt{π}}) m$

Electrical resistance $= 10 Ω$

Electric field strength $= 10 V m^{- 1}$

Current density, $J = \frac{I}{A} = \frac{V}{R A} = \frac{E L}{R A}$

$\Rightarrow J = \frac{10 \times 10}{10 \times π {(\frac{10^{- 2}}{\sqrt{π}})}^{2}} = 10^{5} A m^{- 2}$

Stream:NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: PYQ_2022

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