establish the formula of intensity of magnetic field at the centre of a current carrying coil
Answers
Answer:Consider a circular coil of radius r with n turns carrying i. The intensity of magnetic field at its centre O due to an small part AB=\Delta l=Δl will be.
AB=\displaystyle\frac{\mu_o}{4\pi}\frac{i\Delta l sin 90^o}{r^3}AB=
4π
μ
o
r
3
iΔlsin90
o
[Here \theta=90^oθ=90
o
because the radius is normal to the circumference.]
The intensity of magnetic field at the centre O due to the entire coil,
\sum AB=\sum\displaystyle\frac{\mu_o i\Delta l}{4\pi r^2}∑AB=∑
4πr
2
μ
o
iΔl
[\because sin90^o=1][∵sin90
o
=1]
B=\displaystyle\frac{\mu_o i}{4\pi r^2}\sum \Delta l=\frac{\mu_o}{4\pi}\frac{i}{r^2}\times 2\pi rnB=
4πr
2
μ
o
i
∑Δl=
4π
μ
o
r
2
i
×2πrn
Here \sum \Delta B=B∑ΔB=B and \sum \Delta l=2\pi rn=∑Δl=2πrn=circumference of the coil
\therefore B=\displaystyle\frac{\mu_o}{4\pi}\frac{2\pi ni}{r}∴B=
4π
μ
o
r
2πni
B=\displaystyle\frac{\mu_o ni}{2r}N/AmB=
2r
μ
o
ni
N/Am
This is the required expression.
If the direction of current in the coil is anticlockwise, the direction of magnetic field will be normal to the plane of paper upwards and if current in the coil is clockwise, the direction of magnetic field is normal to the plane to paper downwards.
Explanation: