Question

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. 

Answer 1

According to Biot-Savart’s law, magnetic field due to a current element is given by 

vector dB = (μ₀ Idl x r)/(4πr²) where r = √(x² + a²) 

∴  dB = (μ₀ Idl sin 90°)/(4π)(x² + c²)

And direction of vector dB is ⊥ to the plane containing I vector dl and vector r. 

 Resolving vector dB along the x – axis and y – axis. 

 dB_x = dB sin θ 

 dB_y = dB cos θ 

taking the contribution of whole current loop we get 

B_x = ∮dB_x = ∮dBsin θ =

For centre x = 0 

∴ |vector B₀| = (μ₀ 2Iπa²)/ (4πa³)  = μ₀(1/2a) in the direction of vector m