Suppose a point P is at a distance a from a straight, infinitely long, wire carrying a current I, as shown in below figure. The incremental magnetic induction \(d\vec B\) at he point P due to a current element, I \(\vec{dl}\) is
At the point P, \(\vec{dB}\) is directed perpendicular to the plane of the figure and into of the page as given by the right hand rule for the direction of \(\vec{dl}\) x \(\hat r\).
At point P, \(\vec{dB}\) has this same direction for all the current elements into which the wire can be divided. Thus, we can find the magnitude of the magnetic field produced at P by the current elements in the lower half of the infinitely long wire by integrating dB in Eq. (2),
from 0 to ∞.
magnetic induction in the vicinity of a current in a straight infinitely long wire
Now consider a current element in the upper half of the wire, one that is as far above P as I \(\vec{dl}\) is below P. By symmetry, the magnetic field produced at P by this current element has the same magnitude and direction as that from I \(\vec{dl}\) in above figure. Thus, the magnetic field produced by the upper half of the wire is exactly the same as that produced by the lower half. Hence, the magnitude of the total magnetic field at P is
From r = \(\sqrt{l^2+a^2}\)
\(\therefore\) sin \(\theta\) = \(\cfrac ar\) = \(\cfrac{a}{\sqrt{l^2+a^2}}\)
That is, the magnitude B is inversely proportional to the distance from the wire. Because of the axial symmetry about the straight wire, the magnetic induction has the same magnitude B at all points on a circle in a transverse plane and centred on the conductor; the direction of \(\vec B\) is everywhere tangential to such a circle. Thus, the magnetic field lines around the current in the straight wire is a family of circles centred on the wire.
[Notes : (1) The magnetic field at P due to either only the lower half or the upper half of the infinite wire in figure is half the value in Eq. (5); that is, for a semi-infinite wire,