The given figure shows the cross-section of a long straight solid cylindrical wire of radius R carrying a steady current I uniformly distributed across its cross-section. X and Y are two points located at radial distances r_X = R 4 and r_Y = 2R from the axis of the wire. The ratio of the magnetic field magnitudes B_X : B_Y is:

Step-by-step solution

For a long straight solid cylindrical wire of radius R carrying a steady current I uniformly distributed across its cross-section, the magnetic field at a radial distance r is given by Ampere's law: Inside the wire ( r R ): B_ in = ₀ I r 2 R^2 Outside the wire ( r R ): B_ out = ₀ I 2 r For point X , the radial distance is r_X = R 4 (inside the wire): B_X = ₀ I ( R 4 ) 2 R^2 = ₀ I 8 R For point Y , the radial distance is r_Y = 2R (outside the wire): B_Y = ₀ I 2 (2R) = ₀ I 4 R The ratio of the magnetic field magnitud