Write the expression for the magnetic induction at any point on the axis of a circular current carrying coil. Hence, obtain an expression for the magnetic induction at the centre of the circular coil ? (Please answer ASAP and correctly)
Answers
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