Question

. Derive an expression for magnetic field
at the centre of circular current carrying
coil.
​

Answer 1

Answer:

B=μoI/2a

Explanation:

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  dB  at the centre O of the coil due to current element Idl is given by,

dB=μoIdlasinθ/4a²

dB=μoIdlsinθ/4πa2

The direction of dB  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 loopB=

⎰dB=⎰μoIdlsinθB/4πa2

For each current element, angle between  dl and  r  is 90°. Also distance of each current element from the center O is a.

B=μoIsin90o/4paye a²⎰dl

But ⎰dl=2πa=total length of the coil

B=4πa2/μoI 2πa

B=2a/μoI