# 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.

1

2015-09-10T19:22:59+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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 :

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.

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:

click on thanks button above
Answer is 4 times above value… as there are 4 sides of the square and the fields due to each of them add up. B = = √2 μ I / (π a)
sorry for the omission
thanks a lot
click on thanks button ...