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Using Ampere.s circuital law, obtain an ...

Using Ampere.s circuital law, obtain an expression for the magnetic flux density .B. at a point .X. at a perpendicular distance .r. from a long current carrying conductor. (Statement of the law is not required).

Text Solution

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The figure shows a circular loop of radius r around an infinitely long straight wire carrying current I.
As the field lines are circular, the field `vecB` at any point of the circular loop is directed along the tangent to the circle at that point. By symmetry, the magnitude of field `vecB` is same at every point of the circular loop.

Therefore, `ointvecB.vec(dl)=ointBdlcos0^@=Bointdl =B.2pir`
From Ampere.s circuital law,
`ointvecB.vec(dl)=mu_0I`
`implies B.2pir=u_0I`
`:. B=(mu_0I)/(2pir)`
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