Question

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

Answer 1

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 :

$B=\frac{\mu_0\ I}{4\ \pi\ d}\ [Sin\ \alpha+\ sin\ \beta ],\ \ \ \ ..(1)$
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.
$\vec{dB}=\frac{\mu_0}{4\pi}\frac{I\ \vec{dL} \times \vec{r'}}{|r'|^3},\ \ \ \ ..(2)$
   where r' is the relative vector from the small element dl carrying current I to the point P at which dB is calculated.

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)

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This expression (3) can also be derived by using the Ampere's circuital law:
$\int\limits^{}_{} {\vec{B}.\vec{dL} } \, {}=\mu_0\ I$
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,
$\mu_0*I=\int\limits^{}_{} {\vec{B}.\vec{dL}} \, {}= B * 2\pi*d\\\\So\ \boxed{\ B=\frac{\mu_0 I}{2\pi d}}$

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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:

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