A long straight solid cylindrical wire of radius a carries a steady current. The current density J across the cross-section of the wire is not uniform; it varies with the radial distance r from the axis as J(r) = J₀ r , where J₀ is a positive constant. The radial distance r ( r < a ) at which the magnetic field inside the wire is half of its magnitude at the surface of the wire is:

Step-by-step solution

To find the magnetic field inside the wire at a distance r from the axis, we first calculate the current enclosed by a circular loop of radius r : I_ encl = ₀^r J(r') 2 r' dr' I_ encl = ₀^r (J₀ r') 2 r' dr' = 2 J₀ ₀^r r'^2 dr' = 2 J₀ r^3 3 Applying Ampere's circuital law, B d l = ₀ I_ encl : B(r) 2 r = ₀ ( 2 J₀ r^3 3 ) B(r) = ₀ J₀ r^2 3 The magnetic field at the surface of the wire ( r = a ) is: B(a) = ₀ J₀ a^2 3 We are given that the magnetic field at distance r is half of the magnetic field at the surface: B(r) =