The current density in a cylindrical wire of radius 4 mm is 4 × 1 0 6 A m − 2 . The current through the outer portion of the wire between radial distances 2 R and R is _____ π A .

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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicMedium2 minQB

PhysicsMediumnumerical

The current density in a cylindrical wire of radius $4 mm$ is $4 \times 1 0^{6} A m^{- 2}$ . The current through the outer portion of the wire between radial distances $\frac{R}{2}$ and $R$ is _____ $π A$ .

Answer:

48.00

Solution:

Current density is defined as current per unit cross-section area. Therefore,

$i = J A = J \times π (R^{2} - \frac{R ^{2}}{4})$

$\Rightarrow i = 4 \times 1 0^{6} \times π \times 16 \times 1 0^{- 6} \times \frac{3}{4}$

$\Rightarrow i = 48 π A$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: QB

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