Question

magnetic field due to straight current carrying conductor

Answer 1

Magnetic field due to straight current carrying conductor
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Consider an infinitely long conductor AB through which current I flows. Let P be any point at a distance a from the centre of conductor. Consider dl be the small current carrying element at point c at a distance r from point p. α be the angle between r and dl. l be the distance between centre of the coil and elementary length dl.
From biot-savart law, magnetic field due to current carrying element dl at point P is
from above three equations
Total magnetic field due to straight current carrying conductor is
This is the final expression for total magnetic field due to staright current carrying conductor.
If the conductor having infinite length then,
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Answer 2

Consider a straight conductor XYXY carrying current II. We wish to find its magnetic field at the point PP whose perpendicular distance from the wire is a i.e. PQ=aPQ=a 
Consider a small current element dldl of the conductor at OO. Its distance from QQ is II. i.e OQ=IOQ=I. Let r→r→ be the position vector of the point PP relative to the current element and θθ be the angle between dldl and r→r→. According to Biot savart law, the magnitude of the field dBdB due to the current element dldl will be 

           dB=μ04π.Idlsinθr2dB=μ04π.Idlsin⁡θr2 

From right ΔOQP,ΔOQP, 
         θ+ϕ=900θ+ϕ=900 
    or             θ=900−ϕθ=900−ϕ 
∴sinθ=(900−ϕ)=cosϕ∴sin⁡θ=(900−ϕ)=cos⁡ϕ 
Also         cosϕ=arcos⁡ϕ=ar 
    or                   r=acosϕ=asecϕr=acos⁡ϕ=asec⁡ϕ 
As            tanϕ=latan⁡ϕ=la    
∴l=atanθ∴l=atan⁡θ 
On differentiating, we get  
                       dl=asec2ϕdϕdl=asec2⁡ϕdϕ 
Hence             dB=μ04πI(asec2ϕdϕ)cosϕa2sec2ϕdB=μ04πI(asec2⁡ϕdϕ)cos⁡ϕa2sec2⁡ϕ 
or                  dB=μ0I4πacosϕdϕdB=μ0I4πacos⁡ϕdϕ 

According to right hand rule, the direction of the magnetic field at the PP due to all such elements will be in the same direction, namely; normally into the plane of paper. Hence the total field B→B→ at the point PP due to the entire conductor is obtained by integrating the above equation within the limits −ϕ1−ϕ1 and ϕ2.ϕ2. 

   B=∫ϕ2−ϕ1dB=μ0I4πa∫ϕ2ϕ1cosϕdϕB=∫−ϕ1ϕ2dB=μ0I4πa∫ϕ1ϕ2cos⁡ϕdϕ 
  =μ0I4πa[sinϕ]ϕ2−ϕ1=μ0I4πa[sin⁡ϕ]−ϕ1ϕ2 
 =μ0I4πa=μ0I4πa [sinϕ2−sin(−ϕ1)][sin⁡ϕ2−sin⁡(−ϕ1)] 
       or 
 B=μ0I4πaB=μ0I4πa [sinϕ2+sinϕ1]