Question

A flat circular coil which has 𝑁 turns each of diameter d carries a current 𝐼. If 𝜇0 is the

permeability of free space, what is the expression of magnetic field 𝐵 at the center of the

coil?

Answer 1

The intensity of the magnetic field at any point is obtained by the Biot- Savart's law.

This law in vector form can be written as

$\overrightarrow {dB} = \dfrac{\mu_0}{4 \pi}i \dfrac{\overrightarrow dl\times \vec r}{r^2}$

Now, consider a current carrying circular loop having its centre at O carrying current i.

If dl is a small element at a distance r then, the magnetic field intensity on that point can be written using Biot- Savart's law.

If the coil has N number of turns.

$\overrightarrow {dB} = N\dfrac{\mu_0}{4 \pi}i \dfrac{\overrightarrow {dl}\times \vec r}{r^2}$

$|\overrightarrow {dB}| =N \dfrac{\mu_0}{4 \pi}i \dfrac{idlsin\theta}{r^2}$

As the loop is circular then,

$\theta = 90^\circ$

$sin\theta = 1$

Putting this in the above equation we get,

$|\overrightarrow {dB}| = N\dfrac{\mu_0}{4 \pi}i \dfrac{idl}{r^2}$

The circular loop is composed of numbers of such small elements dl, the we will get the magnetic intensity over the loop. So to get the total field we must sum up that is integrate the magnetic field all over the field,

$B = \int _{0}^{B} \overrightarrow {dB}$

$B = \int \dfrac{N\mu_0}{4\pi} \dfrac{idl}{r^2}\\$

$B = \dfrac{N\mu_0i}{4\pi r^2}\int dl$

The integration of dl gives circumference of loop, we can write

$\int dl = 2\pi r$

$B = \dfrac{N\mu_0i}{4 \pi r^2} 2\pi r$

$B = \dfrac{N\mu_0i}{2r}$