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# A long, straight wire of radius R carries a current distributed uniformly over its cross section. T he magnitude of the magnetic field is (a) maximum at the axis of the wire (b) minimum at the axis of the wire (c) maximum at the surface of the wire (d) minimum at the surface of the wire.

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## (b) minimum at the axis of the wire (c) maximum at the surface of the wire A long, straight wire of radius R is carrying current i, which is uniformly distributed over its cross section. According to Ampere's law, $\oint \stackrel{\to }{B}.d\stackrel{\to }{l}={\mathrm{\mu }}_{\mathrm{o}}{i}_{\mathrm{inside}}\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{surface},\phantom{\rule{0ex}{0ex}}B×2\mathrm{\pi }R={\mathrm{\mu }}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}⇒{B}_{\mathrm{surface}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}i}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}\mathrm{Inside},B×2\mathrm{\pi }r={\mathrm{\mu }}_{\mathrm{o}}i\text{'}\mathrm{for}r Here i, is the current enclosed by the amperian loop drawn inside the wire. Binside will be proportional to the distance from the axis. On the axis B =0 The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre. $\mathrm{Outside},B×2\mathrm{\pi }r={\mathrm{\mu }}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}⇒{B}_{\mathrm{outside}}=\frac{{\mathrm{\mu }}_{\mathrm{o}}i}{2\mathrm{\pi }r},r>R$ Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.  Suggest Corrections  0      Similar questions  Related Videos   Applications of Ampere's Law
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