Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.

Explanation

Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point 0. Consider an element of length dl of the wire at a distance I from point O and `vecr` be the vector joining the element dl with the point P. Let `theta` be the.angle between dl and `vecr` . Then, the magnetic field at P due to the element is
`dvecB = (mu_(0)Idvecl)/(4pir^(2)) sin theta` (Unit vector perpendicular to `vecdl` and `vecr`) .........(i)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors and `vecdl` (let it be `hatn`). The net magnetic field can be determined by integrating equation with proper limits.
`vecB = int dvecB`

From the figure, in a right angle triangle PAO,
`tan(pi - theta) =a/l`
`l =-a/(sin theta)` ,since `tan(pi - theta) =- tan theta rArr 1/(tan theta) = cot theta`
`l=-a cot theta` and `r = a "cosec"theta`
Differentiating,
`dl = a "cosec"^(2) theta, d theta`
`dvecB =(mu_(0)I)/(4pi) (a "cosec"^(2)theta d theta)/(a " cosec "theta)^(2) sin d theta hat`
`dvecB =(mu_(0)I)/(4 pi a) sin theta d theta hatn`
This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. 0. Therefore, the net magnetic field at the point P which can be obtained by integrating `dvecB` by varying angle from `theta = varphi_(1)` to `theta = varphi_(2)` is
`vecB =(mu_(0)I)/(4pi a) int_(varphi_(1))^(varphi_(2)) sin theta d theta hat n = (mu_(0)I)/(4 pi a) (cos varphi_(1) - cos varphi_(2))hatn`
Note that here `hatn` represents the unit vector from the point O to P.