# Find the magnetic field exerted by a long wire at a point 2 m away from it.

by agrawaldevansh 10.09.2015

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by agrawaldevansh 10.09.2015

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

Here α and β are the angles APB and BPA.

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.

Now for a very long conductor with L >> d, B = μ₀ I / 4 π d [ sin 90 + sin 90 ]

* B = **μ₀ I / 2 π d * --- (3)

======================

This expression (3) can also be derived by using the Ampere's circuital law:

The line integral has the dot product of magnetic field and the path length vector at each point on a closed path in space, that encloses a net outgoing current I.

Let us take a circular path (2-d circle) of radius d around the long conductor. The magnetic field B at each point on the circumference is the same from symmetry. B points tangential direction. Hence the angle between dL and B is 0. Thus,

===============

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:

Here α and β are the angles APB and BPA.

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

Now for a very long conductor with L >> d, B = μ₀ I / 4 π d [ sin 90 + sin 90 ]

======================

This expression (3) can also be derived by using the Ampere's circuital law:

The line integral has the dot product of magnetic field and the path length vector at each point on a closed path in space, that encloses a net outgoing current I.

Let us take a circular path (2-d circle) of radius d around the long conductor. The magnetic field B at each point on the circumference is the same from symmetry. B points tangential direction. Hence the angle between dL and B is 0. Thus,

===============

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: