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Question
A long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a.
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Solution
`ointvecB*vecdl = mu_0I_(enclosed)`
`(I_(enclosed))/(pia^2) = I/(pir^2)`
`I_(enclosed) = I r_2/a^2`
`vecB*vecdl = Bdl (because costheta = 1)`
`therefore ointBdl = mu_0Ir_2/a^2`
`Bointdl = mu_0Ir^2/a^2`
`B(2pir) = mu_0 Ir^2/a^2`
`B = (mu_0)/(2pi) I/a^2 r `
(ii) For r > a
From Ampere’s circuital law,
`ointvecB*vecdl = mu_0 l_(enclosed)`
`vecB*vecdl =Bdt cos theta`
`theta = 0`
`I_(enclosed) = I`
`oint Bdl = mu_0I`
`Bointdl = mu_0I`
`B(2pir) = mu_0I`
`B= (mu_0)/(2pi) I/r`
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