A long straight wire of radius `a` carries a steady current `i`. The current is uniformly distributed across its cross section. The ratio of the magnetis field at `(a)//(2) and (2a)` is

Options

A. `1//4`
B. `4`
C. 1
D. `1//2`

Answer

C

Explanation

Current density `J=I/(pia^(2))`
From Ampere's circuital law `ointB.dI=mu_(0).I_("enclosed")`
For `rlta`
`Bxx2pir=mu_(0) 5 J xxpir^(2)impliesB=(mu_(0)I)/(pia^(2))xxr/2`

At `r=a/2,B_(1)=(mu_(0)!)/(4pia^(2))xxr/2`
For `rgta,Bxx2pir=mu_(0)IimpliesB=(mu_(0)I)/(2pir)`
At `r=2a, B_(2)=(mu_(0)I)/(4pia)" ":." "(B_(1))/(B_(2))=1`