Question

Using Biot − Savart’s law, derive the expression for the magnetic field in the vector form at a point on the axis of a circular current loop?

Answer 1

Magnetic field on the axis of a circular current loop

I →Current in the loop, R →Radii of the loop, X-axis →Axis of the loop, x →Distance of OP

dl →Conducting element of the loop

According to Biot-Savart’s law, the magnetic field at P is `dB =(μ_0I|dI xx r)/(4pir^3)`

r² = x²+ R²

| dl × r | = r dl (∵ they are perpendicular)

`thereforedB = (μ_0)/(4pi) (IdI)/((x^2 +R^2))`

dB has two components − dB_x and dB_⊥. dB_⊥ is cancelled out and only the x-component remains.

∴ dB_x= dBcosθ

`Cos theta = R/((x^2 + R^2)^(1/2)`

`therefore dB_x =(μ_0IdI)/(4pi) R/(x^2 +R^2)^(3/2)`

Summation of dl over the loop is given by 2πR.

`B =B_xhati =(μ_0IR^2)/(2(x^2 + R^2)^(3/2 ) )hati`