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

Explanation

Since uniform current is flowing through a straight wire, the current enclosed by the first amperian path is
`i=(Ipir_1^2)/(piR^2)=(Ir_1^2)/(R^2)`
Using, Ampere circuital law,
`oint B* dvecl=mu_0i`
`:. B 2pir_1=mu_0i`
or `B=(mu_0i)/(2pir_1)=(mu_0)/(2pir_1)xx(Ir_1^2)/(R^2)=(mu_0Ir_1)/(2piR^2)`
Magnetic field induction at a distance `r_2` is
`B^'=(mu_0)/(4pi)(2I)/(r_2) :. B/B^'=(r_1r_2)/(R^2)=((a//2)2a)/(a^2)=1`