Question

Using Biot-Savart law, show that magnetic flux density 'B' at the centre of a current carrying circular coil of radius R is given by: `B = ("μ"_0I)/(2R)` where the terms have their usual meaning.

Answer 1

Let a circular coil of radius R carry a current I and O be its centre.

By Biot-Savart's law, the magnitude of the magnetic field at O due a small element δ l of the loop is,

`δB = "μ"_0/(4π) (Iδl sinθ)/R^2`

Where θ is the angle between the length of the element (δ l) and the line joining the element to the point O. Here, θ = 90° because every element of. a circle is perpendicular to the radius.

∴ `δB = "μ"_0/(4π) (Iδl)/R^2`

Every component of the coil will have the same field direction. Because of the complete coil, the field `vec B` magnitude at O is:

`B = "μ"_0/(4pi)I/R^2 "∑"δl`

But `"∑ "δl = 2piR`

∴ `B = "μ"_0/(4pi) · I/R^2 xx 2piR`

or `B = ("μ"_0I)/(2R)`