Consider a long straight wire NM with current I flowing from N to M as shown in Fig. Let P be the point at a distance a from point . Consider an element of length dl of the wire at a distance l from point O and `hat r` be the vector joining the element dl with the point P. Let `theta` be the angle between `vec(dl)" and " hatr`. Then , the magnetic field at P due to the element is `vec(dB) = (mu_(0)I)/(4pi) (vec(dl))/r^(2) sin theta (" unit vector perpendicular to " vec(dl) and vecr)`
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 `vec(dl)" and " hat r ` ( let it be `hat n`) . The net magnetic field can be determined by integrating equation with proper limits .
From the Figure, in a right angle triangle PAO,
`tan ( pi - theta) = a/l`
` l = - a/ (tan theta) ( "since "tan(pi - theta) = - tan theta)`
` l = - a cot theta and r = a cosec theta`
Differentiating ,
`dl = a cosec .^(2) theta d theta`
`vec(dB) = ( mu_(0)I)/(4 pi) ((a cosec . ^(2) theta d theta))/ ((a cosec theta )^(2)) sin theta hat n `
` vec(dB) = (mu_(0)I)/(4 pi) ((a cosec . ^(2) theta d theta ))/((a^(2) cosec ^(2) theta)) sin theta hatn `
` = (mu_(0)I)/(4 pi a) sin theta d theta hat n`
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. `theta`. Therefore, the net magnetic field at the point P can be obtained by integrating `vec(dB)` by varying the angle from `theta = varphi _(1) " to " theta = varphi_(2) ` is
`vecB = (mu_(0)I)/(4 pi a) underset(varphi_(1))overset(varphi_(2))int sin theta d theta hatn = (mu_(0)I)/(4 pi a) ( cos varphi_(1) - cos varphi_(2)) hat n `
For a an infinitely long straight wire , `varphi_(1) = 0 " and " varphi_(2) = pi ` , the magnetic field is
` vecB = (mu_(0)I)/ (2 pi a) hat n`
Note that here `hat n` represents the unit vector from the point O to P .
