Current Electricity

JEE Main level, numerical answer, medium

19 of 380

Question 19 of 380

The current density in a cylindrical wire of radius 4mm4mm is 4×106Am−24\times {10}^{6}A{m}^{-2}. The current through the outer portion of the wire between radial distances R2\dfrac{R}{2} and RR is _____ πA\pi A.

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Answer: 48

Current density is defined as current per unit cross-section area. Therefore,
i=JA=J×π(R2−R24)i=JA=J\times \pi \left({R}^{2}-\dfrac{{R}^{2}}{4}\right)
⇒i=4×106×π×16×10−6×34\Rightarrow i=4\times {10}^{6}\times \pi \times 16\times {10}^{-6}\times \dfrac{3}{4}
⇒i=48πA\Rightarrow i=48\pi A

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