Deduce the expression for the magnetic field at a point on the axis of a current carrying circular loop of radius ‘R’ distant ‘x’ from the centre. Hence, write the magnetic field at the centre of a loop.
`dB =mu_0/(4pi) (Idlxxr)/r^3`
Here, `mu_0/(4pi)`is a constant of proportionality.
The magnitude of this field is given as
Magnetic field on the axis of a circular current loop:
Consider a circular loop carrying a steady current I. The loop is placed in the y–z plane
with its centre at origin O and has a radius R.
Let x be the distance of point P from the centre of the loop where the magnetic field is to
be calculated. Consider a conducting element dl of the loop. The magnitude dB of the
magnetic field due to dl is given by the Biot–Savart’s law as
From the figure, we see that r2 = x2 + R2.
Any element of the loop will be perpendicular to the displacement vector from the element to the axial point. Hence, we have `|dlxxr|=rdl` Thus, we have
`dB=mu_0/(4pi) (Idl)/r^2=mu_0/(4pi) (Idl)/(x^2+R^2) ".....(1)"`
The direction of dB is perpendicular to the plane formed by dl and r. It has an xcomponent
dBx and a component perpendicular to x-axis dB⊥
The perpendicular components cancel each other when summed over. Therefore, only the x component contributes. The net contribution is obtained by integrating dBx = dB cosθ
From the figure, we see that
From equations (1) and (2), we get
The summation of dl yields circumference of the loop 2πR. Hence, the magnetic field at
point P caused by the entire loop is
Case: At the centre of the loop
At the centre x = o, so we have