Current Electricity

JEE Main level, numerical answer, medium

18 of 380

Question 18 of 380

The current density in a cylindrical wire of radius r=4.0mmr=4.0mm is 1.0×106A2m21.0\times {10}^{6}{A}^{2}{m}^{2}. The current through the outer portion of the wire between radial distances r2\dfrac{r}{2} and rr is xπAx\pi A; where xx is

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Answer: 12

Current =J×=J\times Area
=J×π[R2−(R2)]2=Jπ3R24=106×π×3×(4×10−34)2=J\times \pi {\left[{R}^{2}-\left(\dfrac{R}{2}\right)\right]}^{2} =J\pi \dfrac{3{R}^{2}}{4} ={10}^{6}\times \pi \times 3\times {\left(\dfrac{4\times {10}^{-3}}{4}\right)}^{2}
⇒π×3×164=12πA\Rightarrow \pi \times 3\times \dfrac{16}{4}=12\pi A
Hence, the value of x=12x=12.

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