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Derive an expression for magnetic field ...

Derive an expression for magnetic field due to straight long wire carrying current, using Biot-Savart's law at a point 'a' metre away from it.

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Consider a long straight conductor XY, lying in the plane of paper and carrying current I from X to Y. Let P be the point at perpendicular distance a from the straight conductor such that PC = a. Consider a current element of length `vec (dl)` at O. Let `vec r` be the position vector of point P with respect to `vec (dl)` and be the angle between `vec (dl) and vec F.`
According to Biot-Savart.s law, magnetic field induction dB at point P, due to current element `vec (dl)` is given by
`d B = ( mu _(0))/( 4pi ) . ( I dl sintheta )/( r ^(2)) " "...(i)`
From right angled `Delta PCO`
`theta + phi = 90^(@)`
`implies theta = (90^(@) - phi )`
`implies sin theta = sin ( 90^(@) - phi)`
`implies sin theta = cos phi " "...(ii)`
Also from right angled `Delta PCO`
`(a)/(r) = cos phi implies (r)/(a) = (1)/( cos phi ) " "...(iii)`
Again from right angle `Delta PCO`
`(l)/(a) = tan phi implies l = a tan phi`
Differentiating both sides w.r.t. `phi`
`(dl )/(d phi ) = (d)/(d phi) (a tan phi) implies dl = a sec ^(2) phi d phi `

`implies dl = (a)/(c os ^(2) phi) d phi " "...(iv)`
Putting equations, (ii),(iii), (iv) in (i), we get ,
`d B = ( mu _(0))/( 4pi) (I ((a )/( cos ^(2) phi ) dphi ) cos phi )/((a ^(2))/( cos ^(2) phi )) implies d B = ( mu _(0))/( 4pi) .(I cos phi d phi )/(a ) .....(v)`
The direction of dB will be perpendicular to the plance of paper and is directed inwards. Integrating equation (v) between limits `- phi _(1) ` to `phi _(2)`
`int d b = int _(- phi _(1)) ^( + phi _(2)) ( mu_(0))/( 4pi) . (I cos phi d phi )/( a ) implies B = ( mu _(0) I )/( 4pi a ) int _( - phi _(1)) ^( + phi _(1)) cos phi d phi `
`implies B = ( mu _(0) I )/( 4pi a ) [ sin phi ] _( - phi _(1)) ^( + phi _(2)) implies B = ( mu _(0) I )/( 4 pi a ) [ sin phi _(2) - sin ( - phi _(1)) ]`
`implies B = ( mu _(0) I )/( 4pi a ) [ sin phi _(2) + sin phi _(1) ]" "....(vi)`
If the wire is infinitely long. Then, `phi _(1) = phi _(2) = 90^(@)`
`therefore ` equation (vi) become
`B = ( mu _(0) I )/( 4 pi a ) [ sin 90^(@) + sin 90^(@)] = ( mu _(0) I )/( 4 pi a ) [1 + 1] = ( mu _(0) I )/( 2pi a )`
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