(c) Required arrangement is shown in figure. According to Ampere’s circuital law
$B_{out} = \frac{m_{0}}{4p}\frac{2I}{r}$
[image:4_image99.emf]
For an internal point, $r < R$
$B_{internal} = \frac{\mu_{0}(0)}{2\pi r} = 0$
For a point on the pipe, $r = R$
$B = \frac{\mu_{0}I}{2\pi r}$
For an external point, $r < R$
$B_{external} = \frac{\mu_{0}I}{2\pi r}$
Therefore, option (c) is correct.