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A wire of radius 0*8cm carries a current...

A wire of radius `0*8cm` carries a current of `100A` which is uniformly distributed over its cross-section. Find the magnetic field (a) at `0*2cm` from the axis of the wire (b) at the surface of the wire and (c) at a point outside the wire `0*4cm` from the surface of the wire. Neglect the permeability of the material of wire.

Text Solution

Verified by Experts

Here, `R=0*8cm=8xx10^-3m`,
`I=100A`
(a) `B_(i nside)=(mu_0)/(4pi)(2Ir)/(R^2)`
`=10^-7xx(2xx100xx(0*2xx10^-2))/((8xx10^-3)^2)`
`=6*25xx10^-4T`
(b) `B_(surface)=(mu_0)/(4pi)(2I)/(R)=10^-7xx(2xx100)/(8xx10^-3)`
`=2*5xx10^-5T`
(c) When point is outside the wire,
`r=0*8+0*4=1*2cm=1*2xx10^-2m`.
`B_(outside)=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx100)/(1*2xx10^-2)`
`=1*67xx10^-5T`
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