Question

Explain magnetic field due to the current carrying conductor

Answer 1

Answer:

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HERE IS UR ANS.

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][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.