Byampere′slawInsidethewire
∫B.dl=μ0(ienclosed)
∫B.dlcos0=μ0(πa2I.π(2a)2)
B∫dl=μ04I
B(2π(2a))=4μ0I
B=4πaμ0I
Outside the wire,
B′=2πrμ0I=2π(2a)μ0I=4πaμ0I
So, B′B=1.
A long straight wire of radius a carries a steady current I. the current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B’, at radial distances 2a and 2a respectively, from the axis of the wire is:
Held on 30 Apr 2016 · Verified 9 Jul 2026.
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