Question

3. Find the magnetic field induction at a point on
the axis of a circular coil carrying current and
hence find the magnetic field at the centre of
circular coil carrying current. [Art. 4.8]​

Answer 1

Answer:

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 I

dl

is given by,

dB

=

4πr

3

μ

o

I(

dl

×

r

)

where

r

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

dB

at the centre O is

dB=

4πa

3

μ

o

Idlasinθ

∴dB=

4πa

2

μ

o

Idlsinθ

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 loop i.e.

∴B=⎰dB=⎰

4πa

2

μ

o

Idlsinθ

For each current element, angle between

dl

and

r

is 90°. Also distance of each current element from the center O is a.

∴B=

4πa

2

μ

o

Isin90

o

⎰dl

But ⎰dl=2πa=total length of the coil

∴B=

4πa

2

μ

o

I

2πa

∴B=

2a

μ

o

I