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 magnetic field at a /2and 2a is

The situation is shown in fig.Using Ampere's circuital law2π⋅a2(B)a/2=μ0×J×πa22………..(1)and (2π.2a)(B)2a=μ0×J×π(2a)2………(2)Dividing eq. (1) by eq. [2), we get14(B)a/2(B)2a=116(B)a/2(B)2a=14