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← JEE Main 2022 · 27 Jun · Shift 2
PhysicsInteger+4 / -1Question 56 of 90

JEE Main 2022 · 27 Jun · Shift 2, Physics question 56

The current density in a cylindrical wire of radius r=4.0r=4.0 mm is 1.0×1061.0\times10^6 A/m2^2. The current through the outer portion of the wire between radial distances r2\dfrac r2 and rr is xπx\pi A; where xx is ______.

Answer and worked solution

Official answer: 12

Current in the outer shell. Jπ(r2−r24)=106×π×34×16×10−6=12πJ\pi\left(r^2-\frac{r^2}4\right)=10^6\times\pi\times\frac34\times16\times10^{-6}=12\pi A.

Answer: 12