JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Magnetic Effects of Current

Inside the hollow cavity ($r<R/2$), $B=0$ since there is no current enclosed. In the conducting region ($R/2<r<R$), the magnetic field increases from zero at $r=R/2$ following $B=\frac{{\mu }_{0}J}{2\pi r(b^2-a^2)}(r^2-a^2)$. Outside the cylinder ($r>R$), $B=\frac{{\mu }_{0}I}{2\pi r}$, which decreases as $1/r$. Option (D) correctly shows $B=0$ for $r<R/2$, increasing in the conductor, and then decreasing as $1/r$ outside.