Question

Establish the formula for intensity of magnetic field at the center of a current carrying circular coil.

Answer 1

Hint: We will be using the Biot-Savart’s law to find the magnetic field due to infinitesimal current elements along the coil and add up all the results due to all the elements. The sum of the results of all the infinitesimal elements will give the total magnetic field due to the coil.

Formula used:
Biot-Savart’s law
$$d\overrightarrow{B}={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{I|d\overrightarrow{l}\times \overrightarrow{r}|}{r^3}}$$

Complete step by step answer:
We will use the Biot-Savart’s law for that we will need to calculate the value of the magnitude of the cross product of the infinitesimal length vector and the vector r, which is the vector from the center to the infinitesimal element. As for a circle they will always be perpendicular to each other so the magnitude of their cross product will simply be equal to the arithmetic product of both their magnitudes. And the direction of the magnetic field will be given by the right-hand thumb rule. But here we are only concerned with the intensity of the magnetic field.
So now, the equation becomes
$$d\overrightarrow{B}={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Idl\times r}{r^3}}$$
In this equation r is the radius of the loop, dl is the length of the infinitesimal component of the loop, I is the current flowing through the loop, and the rest are physical constants. Now, we will take integral on both sides and we will get the value of the sum of all the effects of infinitesimal parts of the loop.
$$\int d\overrightarrow{B}=\int {\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Idl\times r}{r^3}}={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Ir}{r^3}}\int dl$$
The integral of the magnetic field component will simply be the magnetic field intensity of the loop and the integral of dl will be the total length of the loop i.e. its circumference.

So, the final value of the magnetic field intensity will be
$$B={\displaystyle \frac{{\mu }_o}{4\pi }}{\displaystyle \frac{Ir(2\pi r)}{r^3}}={\displaystyle \frac{{\mu }_{o}I}{2r}}$$


Note:
The question can be solved by a shortcut method. Observe that the magnetic field intensity will be the same for all the infinitesimal elements, then we need not integrate them; we can just replace the length of infinitesimally small elements with the total length of the loop. Here we forgo the integration step and solve the question a little faster.