Magnetic field due to long straight conductor carrying current:Conside a long straight in wire NM with current I flowing from N to M.Let P be the point at a distance a from point O.Consider an element of length dl of the wire at a distance l from point O and `overliner` be the vector joining the element dl with the point P.Let `theta` be the angle between `overline(dl)` and `overliner`,Then ,the magneti field at P due to the elements is
`doverlineB=(mu_0Idoverlinel)/(4pir^(2))sintheta(" unit vector perpendicular to " doverlineland overliner)`
The direction of the fie1ld is perpendicular to the1 plane of the paper and going into it.This can be determined by taking the1 cross product betw1e1e1n to ve1cors `doverline`and `overliner` (Let it be `capn`).The net magnetic field can be determined by intergrating with proper limits.
`overlineB=intd overlineB`
From the figure,ina right angle triangle PAO,`tan(pi-theta)=a/l`
`l=-a/(tan theta)(sin tan(pi-theta)=-tantheta)implies1/(tan theta)=cot theta`
`l=-a cot theta and r=a cosec theta`
Differentiating,
`dl=a cosec^(2) thetad theta`
`doverlineB=(mu_9I(a cosec^(2)thetad theta))/(4pi(a cosectheta)^(2))sinthetad theta hat(n)=(mu_0I(a cosec^(2)thetad theta))/(4pi a^(2)cosec^(2)theta)sinthetad theta hat(n)`
`doverlineB=(mu_0I)/(4pia)sinthetad theta hat(n)`
This is the magetic 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 which can be obtained by intergrating `doverlineB` by varying the angle from `theta=Phi_1 to theta-Phi_2` is
`overlineB=(mu_0I)/(4pira)int_(Phi_1)^(Phi_2)sinthetad theta hat(n)=(mu_0I)/(4pia)(cos Phi_1-cosPhi_2)hat(n)`
For a an infinitely long straight wire ,1=0 and 2=,the magnetic field is `overlineB=(mu_0I)/(2pia)hat(n)....(3)`Note that here `hat(n)` represents the unit vector from the point O to P.
