IIT-JEE 2012 · Paper 2 · Physics Q2

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An infinitely long hollow conducting cylinder with inner radius $R/2$ and outer radius $R$ carries a uniform current density along its length. The magnitude of the magnetic field, $|\vec B|$ as a function of the radial distance $r$ from the axis is best represented by

|B| against r: rises linearly from 0 to R/2, constant from R/2 to R, then decays.

|B| against r: zero up to R/2, jumps to a constant until R, then decays.

|B| against r: zero up to R/2, jumps to a non-zero value and rises until R, then decays.

|B| against r: zero up to R/2, rises continuously from zero to a maximum at R, then decays.

Answer and worked solution

Official answer: D

Ampère's law with current density $J$ in $R/2<r<R$ :

$r<R/2$ : no current enclosed, $B=0$ .
$R/2<r<R$ : $B\cdot2\pi r=\mu_0J\pi\left(r^2-\frac{R^2}4\right)$ , so $B=\dfrac{\mu_0J}{2r}\left(r^2-\frac{R^2}4\right)$ , which starts from zero at $r=R/2$ and rises.
$r>R$ : $B\propto1/r$ .

$B$ is continuous everywhere (no surface currents), so there is no jump at $R/2$ . Graph (D).

(D)

Topics:Ampere's circuital law

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IIT-JEE 2012 · Paper 2, Physics question 2