Using Biot-Savart’s law, derive an expression for magnetic field at any point on axial line of a current carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil.

Explanation

According to Biot-Savart's law, magnetic field due to a current element is given by
`vec(dB) = (mu0)/(4pi) (Ivec(dl)Xbar(x))/(r^(2)) " Where " r = sqrt(x^(2) + a^(2))`
`therefore dB = (mu0)/(4pi) (Idl"sin"90^(@))/(x^(2) + e^(2))`
And direction of `vec(dB) " is " bot` to the plane containing `I vec(dl) " and " vec(r)`.
Resolving `vec(dB)` alonhg the x-axis and y-axis.
`dB_(x) = dB "Sin" theta`
`dB_(y) = dB " cos"theta`
taking the contribution of whole current loop we get
`B_(x) ointB_(x) = ointB"sin"theta = int(mu0)/(4pi) (1dl)/(x^(2) +a^(2))(a)/(sqrt(x&^(2) + a^(2))).`
`B_(x) = (mu0)/(4pi)(1n)/((x^(2) +a^(2))^(E//Z))ointdl = (mu0)/(4pi) (ln xx 2pia)/((x^(2) +a^(2))^(E//Z))`
`" And " B_(y) = ointdB_(y)= ointdB"cos"theta = 0`
`therefore B_(p) = sqrt(B_(x)^(2) +B_(y)^(2)) = B_(x) = (mu0)/(4pi) (2IA)/((x^(2) + a^(2))^(E//Z))`
`therefore vec(B_(p)) = (mu0)/(4pi) (2m)/((x^(2) + a^(2)))(because vec(m) = IA)`
For centre x = 0
`therefore |vec(B_(0))| = (mu0)/(4pi) (2Ipia^(2))/(a^(2)) = mu0((l)/(2a)) " in the direction of "vec(m)`