A long straight hollow cylindrical conductor has an inner radius R and an outer radius 2R . It carries a steady current I that is uniformly distributed over its cross-sectional area. The ratio of the magnitude of the magnetic field at a distance of 1.5 R from the axis to the magnitude of the magnetic field at a distance of 3 R from the axis is :

Step-by-step solution

The cross-sectional area of the hollow cylinder is: A = (2R)^2 - R^2 = 3 R^2 The uniform current density J is: J = I 3 R^2 For a point at a distance r₁ = 1.5 R (which lies inside the conducting material), the area enclosing the current is: A_ encl = (1.5R)^2 - R^2 = (2.25R^2 - R^2) = 1.25 R^2 = 5 4 R^2 The current enclosed by a circular loop of radius 1.5 R is: I_ encl = J A_ encl = ( I 3 R^2 ) ( 5 4 R^2 ) = 5 12 I Applying Ampere's law, the magnetic field B₁ at r₁ = 1.5 R is: B₁ = ₀ I_ encl 2 r₁ = ₀ (5I/12) 2 (1.5