# Find the magnetic field B at the center of square loop of side a carrying a current I

In this post, we will find the formula of the magnetic field **B** at the center of a square loop of side **a, **carrying a current **I**.

Here is the diagram to understand the situation.

**Solution**:

E is the center of the square ABCD. Current I is flowing through ABCD.

the distance between the center E and the midpoint of AB =r = Length of PE=a/2

Now, calculating the magnetic field at center E due to the current carrying AB side, we will use this formula,

B_{AB}=(μ_{0}I)(sinθ1+sinθ2)/(4πr)

B_{AB}=(μ_{0}I)(sin45^{o}+sin45^{o})/(4πa/2)

=>B_{AB}=[(μ_{0} 2 I)/(4πa)](√2)

=>B_{AB}=(μ_{0} 2√2 I)/(4πa) ………….. (1)

Since all sides of the square create the same magnetic field at the center (the direction is also the same), so total magnetic field B will be

B=4×B_{AB}=4×(μ_{0} 2√2 I)/(4πa)

∴B=8√2[(μ

_{0}I)/(4πa)] …………… (2)

Equation (2) can also be expressed as: B=2√2[(μ_{0} I)/(πa)]