CBSE Class 12-science Answered

pl derive the expression for the field due to straight wire using Biot Sawart's law

Asked by Rajeev | 21 Feb, 2016, 11:29: PM

Expert Answer

Consider a straight wire XY lying in the plane of paper carrying current I in the direction X to Y.

Let P be a point at a perpendicular distance a from the straight wire conductor. Clearly, PC = a. Let the conductor be made of small current elements. Consider a small current element

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of the straight wire conductor at O. Let

begin mathsize 14px style straight r with rightwards arrow on top end style

be the position vector of P w.r.t. current element and θ be the angle between

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and

begin mathsize 14px style straight r with rightwards arrow on top end style

. Let CO = $Syntax error from line 1 column 73 to line 1 column 98. Unexpected 'mathvariant'.$ .

According to Biot-Savart's law, the magnetic field

begin mathsize 14px style straight d straight B with rightwards arrow on top end style

(i.e., magnetic flux density or magnetic induction) at point P due to current element

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is given by

begin mathsize 14px style straight d straight B with rightwards arrow on top equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction. fraction numerator straight I space dl with rightwards arrow on top cross times straight r with rightwards arrow on top over denominator straight r cubed end fraction end style

begin mathsize 14px style straight d straight B with rightwards arrow on top equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction. fraction numerator straight I space stack dl space with rightwards arrow on top sin theta over denominator straight r squared end fraction space space space... space left parenthesis 1 right parenthesis end style

In right angled ΔPOC, θ + Φ = 90º or θ = 90º-Φ

Therefore, sin θ = sin (90º-Φ) = cos Φ ... (2)

Also,

begin mathsize 14px style cosϕ equals straight a over straight r space or space space straight r equals fraction numerator straight a over denominator cos space straight ϕ end fraction space space space... space left parenthesis 3 right parenthesis end style

And,

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Differentiating it we get,

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= a sec²ΦdΦ ... (4)

Putting the values of eqn (2), (3) and (4) in eqn (1), we get,

begin mathsize 14px style dB equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction. fraction numerator straight I left parenthesis straight a space sec squared ϕdϕ right parenthesis cosϕ over denominator begin display style fraction numerator straight a squared over denominator cos squared straight ϕ end fraction end style end fraction equals fraction numerator straight mu subscript 0 over denominator 4 straight pi end fraction straight I over straight a cosϕdϕ space space space space... space left parenthesis 5 right parenthesis end style

The direction of

begin mathsize 14px style straight d straight B with rightwards arrow on top end style

, according to the right hand thumb rule, will be perpendicular to the plane of paper and directed inwards. As all the current elements of the conductor will also produce magnetic field in the same direction, therefore, the total magnetic field at point P due to the current through the whole straight wire conductor XY can be obtained by integrating equation (5) within the limits -Φ₁ and +Φ₂.

Thus,