Magnetic Field due to Current Carrying Circular Wire of Infinite Length Using Ampere's law
Consider a circular wire of infinite length and radius $a$ through which a steady current $I$ passes. The current in the wire gives rise to a magnetic field around it. The magnetic field lines are represented by concentric circles with their centres on the axis of the wire.
magnetic field intensity due to the current carrying circular wire of infinite length at the points (i) out side the wire, (ii) on the surface of the wire and (iii) inside the wire.
Case I. Magnetic field intensity at a point outside the wire.
Consider a small portion of finite length of the wire of infinite length.
Let $P_1$ be the point outside the wire at a distance $r$ ($> a$) from the axis of the wire.
According to Ampere's circuital law,
$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I $
OR
$ \oint B dl \cos 0^\circ = \mu_0 I \quad \text{or} \quad B \oint dl = \mu_0 I $
$ B \times 2 \pi r = \mu_0 I $
$ B = \frac{\mu_0 I}{2 \pi r} $
The direction of magnetic field is determined by right hand thumb rule.
Thus, the magnetic field intensity at a point outside the wire varies inversely as the distance of the point from the axis of the wire.
That is,
$ B \propto \frac{1}{r} $
Case II. Magnetic field intensity on the surface of the wire.
In this case, $r = a$ (radius of the wire)
Hence, eqn. (i) can be written as
$B = \frac{\mu_0 I}{2 \pi a} $
Case III. Magnetic field intensity at a point inside the wire.
Consider a point $P_2$ inside the wire at a distance $r$ from the axis of the wire, such that $r < a$
(a) If the current flows only along the surface of the wire, then according to Ampere's circuital law.
$ \oint \vec{B} \cdot d\vec{l} = 0 $
($\because I = 0$ inside the circle of radius $r < a$)
$ B = 0 $
(b) If the current is uniformly distributed throughout the cross-section of the wire, then the current through the circle of radius $r < a$ is given by,
$I^{'}$ = current per unit area of the wire × area of the circle of radius r
$ I' = \frac{I}{\pi a^2} \times \pi r^2 = \frac{I r^2}{a^2} $
According to Ampere's circuital law,
$ \oint \vec{B} \cdot d\vec{l} = \mu_0 I' $
$ \oint B dl \cos 0^\circ = \mu_0 \frac{I r^2}{a^2}$
$B \oint dl = \mu_0 \frac{I r^2}{a^2} $
$ B \times 2 \pi r = \mu_0 \frac{I r^2}{a^2} $
$ B = \frac{\mu_0 I r}{2 \pi a^2} $
Thus,
$ B \propto r $
The variation of magnetic field (B) with distance (r) from the axis of the wire is shown in figure
