# 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.

by LataBadami509 02.09.2015

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by LataBadami509 02.09.2015

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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)*

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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:

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

So now, we have α = π/4 and β = π/4 and d = 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: