Question

State ampere's circuital law, using it derive an expression for magnetic field at a point near a long straight current carrying conductor.

Answer 1

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Consider a straight conductor XY carrying current I. We wish to find its magnetic field at the point P whose perpendicular distance from the wire is a i.e. PQ=a

Consider a small current element dl of the conductor at O. Its distance from Q is I. i.e OQ=I. Let r→ be the position vector of the point P relative to the current element and θ be the angle between dl and r→. According to Biot savart law, the magnitude of the field dB due to the current element dl will be

dB=μ04π.Idlsinθr2

From right ΔOQP,

θ+ϕ=900

or θ=900−ϕ

∴sinθ=(900−ϕ)=cosϕ

Also cosϕ=ar

or r=acosϕ=asecϕ

As tanϕ=la

∴l=atanθ

On differentiating, we get

dl=asec2ϕdϕ

Hence dB=μ04πI(asec2ϕdϕ)cosϕa2sec2ϕ

or dB=μ0I4πacosϕdϕ

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

B=∫ϕ2−ϕ1dB=μ0I4πa∫ϕ2ϕ1cosϕdϕ

=μ0I4πa[sinϕ]ϕ2−ϕ1

=μ0I4πa [sinϕ2−sin(−ϕ1)]

or

B=μ0I4πa [sinϕ2+sinϕ1]

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.

Ampere's Circuital Law states the relationship between the current and the magnetic field created by it. This law says, the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

Hope it helps you