Question

Find an expression for the magnetic field induction at the centre of the coil bent in the form of a square of side 2a and carrying current I.

Answer 1

Let the coil be denoted by ABCD.  Let the center be O.  So, A, B, C and D are 4 corners of the coil.  The distance between the coil and the center is 2a/2 = a.  This is the perpendicular distance between any side and the center O.

The magnetic field strength/intensity B due to a finite conductor carrying a current of I, and of length L at a point at a distance d is :

$B=\frac{\mu_0\ I}{4\ \pi\ d}\ [Sin\ \alpha+\ sin\ \beta ],\ \ \ \ ..(1)$

Here α and β are the angles OAB and OBA.

This expression can be derived from the Biot-Savart's Law and integrating over the length of the conductor.

$\vec{dB}=\frac{\mu_0}{4\pi}\frac{I\ \vec{dl} \times \vec{r'}}{|r'|^3},\ \ \ \ ..(2)$
where r' is the relative vector from the small element dl carrying current I to the point P at which dB is calculated.
 
So now, we have α = π/4  and β = π/4  and  d = a.

B = (μ₀ I / 4 π a) [1/√2 + 1/√2 ] = μ I / (2√2 π a)

=============
The expression (1) can be derived as:

  Let us take an element dy (or dl) at P at a distance y from the center of a conductor AB of length L.

   We want to find magnetic field induction strength B at O, whose perpendicular distance from AB is d.   Let the angle dy (or AB) makes with PO = θ.   Let PO make angle Φ with the perpendicular OC from O onto AB.    θ = π/2 - Φ.    r' = distance PO.

    y = d tanΦ      and so      dy = d sec² Ф dФ
   r' = d / Cos Ф  = d SecФ             

So we get:
$B= \frac{\mu_0\ I}{4 \pi}\int\limits^{-\beta}_{\alpha} {\frac{dy\ Sin\theta}{r'^2}} \, \\\\= \frac{\mu_0\ I}{4 \pi\ d} \int\limits^{-\beta}_{\alpha} {cos\phi} \, d\phi \\\\=\frac{\mu_0\ I}{4\pi\ d}[Sin\alpha+\sin\beta}]$