A long straight wire of a circular cross-section of radius 'a' carries a steady current I. The current is uniformly distributed across the cross-section. Apply Ampere's circuital law to calculate the magnetic field at a point at distance 'r' in the region for (i) `rlta` and (ii) `rgta`.

Explanation

Field at outside points : The Amperean loop is a circle labelled 2 haveing radius `r gt a`.
Length of the loop, `L=2pi r`
Net current enclosed by the loop `=I`
By Ampere's circuital law,
` BL=mu_0I`
or ` Bxx 2pi r =mu_0 I`
or ` B=(mu_0I)/(2pir)`
i.e., `Bprop (1)/(r)` [For outside points]
Field at inside points : The Amperean loop is a circle labelled 1 with `rlt a`.
Length of the loop, `L=2pi r`
So, the current enclosed by loop 1 is less than I. As the current distribution is uniform, the fraction of I enclosed is
` I' =(1)/(pi a^2)xx pir^2=(Ir^2)/(a^2)`
Applying Amphere's law,
` BL= mu_0 I`'
or ` B xx 2pir =mu_0 (lr^2)/(a^2)`
or ` B=((mu_0 I)/(2pia^2))r ["For" r lt a]`
i.e., ` Bprop r` [For inside points ]
Thus, the field B is proportional to r as we move from the axis of the cylinder towards its surface and then it decreases as `(1)/(r)`. The variation of B with distance r from the centre of the wire is shown in figure.