A long straight solid conductor of radius `6cm` carries a current of `8A`, which uniformly distributed over its circular cross-section. Find the magnetic field (a) at a distance of `3cm` from the axis of the conductor (b) at a distance `10cm` from the axis of the conductor.

Answer

(a) `1*33xx10^-5T` (b) `1*6xx10^-5T`

Explanation

Here, `R=6cm=6xx10^-2m`, `I=8A`,
(a) `r=3cm=3xx10^-2m`. Point is inside the conductor
`B=(mu_0)/(2pi)(Ir)/(R^2)=((4pixx10^-7))/(2pi)xx(8xx(3xx10^-2))/((6xx10^-2)^2)`
`=1*33xx10^-5T`
(b) `r=10cm=0*10m`. Point is outside the conductor
`B=(mu_0)/(4pi)(2I)/(r)=10^-7xx(2xx8)/(0*10)=1*6xx10^-5T`