A long straight solid cylindrical wire of radius R carries a steady current which is uniformly distributed across its cross-section. Let B₁ be the magnitude of the magnetic field at a radial distance r = R 2 from the axis of the wire, and B₂ be the magnitude of the magnetic field at a radial distance r = 2R . The ratio B₁ : B₂ is

Step-by-step solution

The magnetic field inside a solid cylindrical wire at a distance r B_ in = ₀ I r 2 R^2 At r = R 2 , the magnetic field is B₁ = ₀ I ( R 2 ) 2 R^2 = ₀ I 4 R The magnetic field outside the wire at a distance r > R is given by B_ out = ₀ I 2 r At r = 2R , the magnetic field is B₂ = ₀ I 2 (2R) = ₀ I 4 R Taking the ratio of B₁ to B₂ , we get B₁ B₂ = ₀ I 4 R ₀ I 4 R = 1 Thus, the ratio is 1:1 . Answer: 1:1