Magnetic field at the centre of current carrying circular coil

Consider a circular coil of radius a and carrying current I in the direction shown in Figure.

Suppose the loop lies in the plane of paper.

It is desired to find the magnetic field at the centre O of the coil. Suppose the entire circular coil is divided into a large number of current elements, each of length dl.

According to Biot-Savart law, the magnetic field

at the centre O of the coil due to current element I

is given by

where

is the position vector of point O from the current element.

The magnitude of at the centre O is

The direction of

is perpendicular to the plane of the coil and is directed inwards.

Since each current element contributes to the magnetic field in the same direction, the total magnetic field B at the center O can be found by integrating the above equation around the loop i.e.

For each current element, angle between and is 90°.

Also distance of each current element from the center O is a.

Therefore

Now,

dl = Total length of the coil

= 2

If the coil has n turns, each carrying current in the same direction, then contributions of all the turns are added up, then B =

Magnetic field on the axis of circular coil carrying current

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 = dB' =

Resolving

and

into rectangular components, it is clear that vertical components (dB cosα and dB' cos α) will be equal and opposite and thus cancel each other.

However, components along the axis of the coil (dB sin α and dB' 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