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Derive an expression for the force between two long parallel current carrying conductors meritnation

Answer 1

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What is the magnetic field due to a long straight current carrying conductor? 

What is the magnetic field due to a long straight current carrying conductor? 



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Consider a straight conductor XYXYcarrying 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 QQis 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 PPdue 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−ϕ1and 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][sin⁡ϕ2+sin⁡ϕ1] 

This equation gives magnetic field due to a finite wire in terms of te angles subtended at the observation point by the ends of wire.

Answer 2

Answer:

Explanation:

elected May 26, 2018 by Vikash Kumar Best answer (a) Suppose two long thin straight conductors (or wires) PQ and RS are placed parallel to each other in vacuum (or air) carrying currents I1 I2 and respectively. It has been observed experimentally that when the currents in the wire are in the same direction, they experience an attractive force (fig. a) and when they carry currents in opposite directions, they experience a repulsive force (fig. b). Let the conductors PQ and RS carry currents I1 I2 and in same direction and placed at separation r.(fig.). Consider a current–element ‘ab’ of length ΔL of wire RS. The magnetic field produced by current-carrying conductor PQ at the location of other wire RS According to Maxwell’s right hand rule or right hand palm rule no. 1, the direction of B1 will be perpendicular to the plane of paper and directed downward. Due to this magnetic field, each element of other wire experiences a force. The direction of current element is perpendicular to the magnetic field; therefore the magnetic force on element ab of length ΔL Read more on Sarthaks.com - https://www.sarthaks.com/57002/derive-an-expression-for-the-force-between-two-long-parallel-current-carrying-conductors