Verified Solution • 2025–26 Session moderate

QJEEM22Pha97a: JEE Main 2022 Physics - Class 12 JEEM22Pha97a Solution: The current density in a cylindrical wire of radius r = 4.0 mm…

The current density in a cylindrical wire of radius r = 4.0 mm is 1.0 ×\times 106 A/m2. The current through the outer portion of the wire between radial distances r2{r \over 2} and r is x π\pi A; where x is __________.

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)24×1.0×106= {{3\pi \times {{(4 \times {{10}^{ - 3}})}^2}} \over 4} \times 1.0 \times {10^6}

=12 π= 12\,\pi

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