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← JEE Main 2022 · 27 Jun · Shift 1
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JEE Main 2022 · 27 Jun · Shift 1, Physics question 56

The current density in a cylindrical wire of radius 4 mm is 4×1064\times10^6 Am−2^{-2}. The current through the outer portion of the wire between radial distances R2\dfrac R2 and R is ______ π\pi A.

Answer and worked solution

Official answer: 48

Current in the outer shell. I=Jπ(R2−R24)=4×106×π×34×16×10−6=48πI=J\pi\left(R^2-\frac{R^2}4\right)=4\times10^6\times\pi\times\frac34\times16\times10^{-6}=48\pi A.

Answer: 48

Topics:Drift velocity and current
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