A long straight solid conductor of radius 5 cm carries a current of 2A, which is uniformly distributed over its circular cross section. Find the magnetic field at a distance of 3 cm from the axis of the conductor.

Explanation

Let us consider an internal point P at a distance `r(=3mc)` from the axis of the conductor. Imagine a circular path or radius r around the conductor, such that P lies on it.
If R is radius of the solid conductor then current enclosed by the circular path,
`I_(1)=(I)/(piR^(2))xxpi r^(2)=(It^(2))/(R^(2))`
Let B be the magnetic field at point P due to the current carrying conductor, B acts tangentially to the conductor. B act tangentially to the circular path . so accroding to ampere's Circuital law,
`oint vecB*d vecl=mu_(0)I_(1)`
`or Bxx 2pi r=(mu_(0)Ur^(2))/(R^(2))`
`or B=(mu_(0))/(pi)*(Ir)/(R^(2))=(mu_(0))/(4pi)*(2Ir)/(R^(2))`
Hence `I=2A,R=3 cm =0.3 m, R=5 cm =0.05 m`
`:. B=(10^(-7)xx2xx2xx0.03)/((0.05)^(2))=4.8xx10^(-6)T`