Verified Solution • 2025–26 Session moderate

QJEEM22Phe57d: JEE Main 2022 Physics - Class 12 JEEM22Phe57d Solution: The current density in a cylindrical wire of radius 4 mm is 4…

The current density in a cylindrical wire of radius 4 mm is 4 ×\times 106 Am −- 2. The current through the outer portion of the wire between radial distances R2{R \over 2} and R is ____________ π\pi A.

Step-by-Step Verified Solution

i=A×ji = A \times j

=π(R2−R24)j= \pi \left( {{R^2} - {{{R^2}} \over 4}} \right)j

=3πR24×j= {{3\pi {R^2}} \over 4} \times j

=3π×(4×10−3)24×4×106= {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 4 \times {10^6}

=48π= 48\pi

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