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Question:

how can we relate angles i.e. phie, of expression : "magnetic field along the axis of a current carrying circular coil

Answer:

Consider a circular coil of radius a, centre O and carrying a current I in the direction shown in figure. Let the plane of the coil be perpendicular to the plane of the paper. It is desired to find the magnetic field at a point P on the axis of the coil such that OP = r. Consider two small current elements, each of length dl, located diametrically opposite to each other at Q and R. Suppose the distance of Q or R from P is x.

i.e. PQ	=	PR = x.
then, x	=
Let
QPO	=	α
	=	RPO

According to Biot-Savart law, the magnitude of magnetic field at P due to current element at Q is given by;

The magnetic field at P due to current element at Q is in the plane of paper and at right angles to

and in the direction shown. Similarly, magnitude of magnetic field at point P due to current element at R is given by;

It also acts in the plane of paper and at right angle to

but in opposite direction to dB. From the above two equations dB = dBdash =

Resolving

and

into rectangular components, it is clear that vertical components (dB cosα and dBdash cos α) will be equal and opposite and thus cancel each other. However, components along the axis of the coil (dB sin α and dBdash sin α) are added and act in the direction PX. This is true for all the diametrically opposite elements of the circular coil. Therefore, when we sum up the contributions of all the current elements of the coil, the perpendicular components will cancel. Hence the resultant magnetic field at point P is the vector sum of all the components dB sin α over the entire coil.

B	=
Now sin α	=	and

If the circular coil has n turns, then,

along PX

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