Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d\bar l\) at C and D Let r be the vector joining the current element (I \(d\bar l\)) at C to the point P.
PC = PD = r = \(r\) = \(\sqrt{R^2 +Z^2}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar l\) is
\(d \vec B\) = \(\frac{μ_0}{4π}\) \(\frac{I \bar dl \times \hat r}{r^2}\) ..............(1)
The magnitude of magnetic field due to current element I \(d\bar l\) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat k\)) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar l\) around the loop \(d \bar B\), sweeps out a cone, then the net magnetic field \(\vec B\) at point P is
Using Pythagorous theorem r2 = R2 + Z2 and integrating line element from 0 to 2πR, we get
Note that the magnetic field \(\vec B\) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.
