magnetic field due to a circular …

Let us consider a circular loop of radius "a" with centre C. Let the plane of the coil be perpendicular to the plane of the paper and "I" be the current flowing in the direction as shown in the figure (coming out from plane of paper normally at top). Suppose P is any point on the axis at a distance "r" from the centre C.

Now, consider a current element "Idl" on top (L) where the current comes out of paper perpendicularly whereas at bottom (M) the current enters into the plane of paper perpendicularly.
{tex}\because \quad L P \perp I d l{/tex}
Also, {tex}M P \perp I d l{/tex}
{tex}\because \quad L P = M P = \sqrt { r ^ { 2 } + a ^ { 2 } }{/tex}

According to Biot-Savart's law,
The magnetic field at point P due to current element Idl
{tex}d B = \frac { \mu _ { 0 } } { 4 \pi } \cdot \frac { I d l \sin 90 ^ { \circ } } { \left( r ^ { 2 } + a ^ { 2 } \right) }{/tex}
where, a = radius of circular loop,
r = distance of point P from centre and along the axis.
The direction of dB is perpendicular to LP and along PQ, so {tex}P Q \perp L P{/tex}.
Similarly, the value of magnetic field at point P due to current element Idl at the bottom is same and its direction is along PQ', where {tex}P Q ^ { \prime } \perp M P{/tex} .
Let ϕ = the angle between LP and CP.
Now, resolving magnetic field dB into two components, (i) dBsinϕ along the axis and (ii) dBcosϕ normally to the axis.
the components dBcosϕ balance each other and axial components are in same direction so get added. So,
Net magnetic field is given by
{tex}B = \oint d B \sin \phi = \oint \frac { \mu _ { 0 } } { 4 \pi } \left( \frac { I d l } { r ^ { 2 } + a ^ { 2 } } \right) \cdot \frac { a } { \sqrt { r ^ { 2 } + a ^ { 2 } } }{/tex} {tex}\left[ {\because In \Delta PCL,\sin \phi = \frac{a}{{\sqrt {{r^2} + {a^2}} }}} \right]{/tex}
or {tex}B = \frac { \mu _ { 0 } I a ^ { 2 } } { 2 \left( r ^ { 2 } + a ^ { 2 } \right) ^ { 3 / 2 } }{/tex}
For N turns, {tex}B = \frac { \mu _ { 0 } N I a ^ { 2 } } { 2 \left( r ^ { 2 } + a ^ { 2 } \right) ^ { 3 / 2 } }{/tex}Tesla.
Magnetic field lines due to a current carrying loop are given by