Question

Using Biot-Savart law, derive expression for the magnetic field `(vecB)` due to a circular current carrying loop at a point on its axis and hence at its centre.

Answer 1

Let us consider a circular loop with radius a and centre C. Let the coil's plane be perpendicular to the plane of the paper, and current flow in the indicated direction. Assume P is any point on the axis at a distance r from the center.

Consider a current element (dl) on top (L) that generally exits the paper and enters the plane at the bottom (M).

LP ⊥ dl

MP ⊥ dl

LP = MP = `sqrt(r^2 + a^2)`

Now, the magnetic field at P due to the current element at L according to Biot-Savart Law,

dB = `mu_0/(4 pi) * (idl sin90^circ)/(r^2 + a^2)`

where, a = radius of circular loop

r = distance of point P from the centre along the axis. dB cos Φ components balance each other, andthe  net magnetic field is given by:

B = `oint dB sinphi`

= `oint mu_0/(4 pi) [(idl)/(r^2 + a^2)] * a/(sqrt(r^2 + a^2))`    ...`[therefore trianglePCM sin phi = a/(sqrt(r^2 + a^2))]`

= `mu_0/(4 pi) (Ia)/((r^2 + a^2)^(3/2)) intdl`

= `mu_0/(4 pi) (Ia)/((r^2 + a^2)^(3//2)) xx 2pi a`

= `(mu_0Ia^2)/(2(r^2 + a^2)^(3/2))`

As the direction of the field is along +ve X-direction, we can write:

`vec B = (mu_0Ia^2)/(2(r^2 + a^2)^(3//2)) hati`

At its centre, r = 0

B = `(mu_0Ia^2)/(2(a^2)^(3/2)`

= `(mu_0I)/(2a)`