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`

Concept: Ampere’s Circuital Law

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