A long straight solid cylindrical wire of radius R carries a steady current I which is uniformly distributed across its cross-section. Let P be a point inside the wire at a radial distance r₁ = R 4 from the axis, and Q be a point outside the wire at a radial distance r₂ = 2R from the axis. The ratio of the magnitude of the magnetic field at point P to that at point Q is:

Step-by-step solution

For a solid cylindrical wire of radius R carrying a uniform current I , the magnetic field at a distance r from the axis is given by: Inside the wire ( r B_ in = ₀ I r 2 R^2 Outside the wire ( r R ): B_ out = ₀ I 2 r At point P ( r₁ = R 4 ): B_P = ₀ I ( R 4 ) 2 R^2 = ₀ I 8 R At point Q ( r₂ = 2R ): B_Q = ₀ I 2 (2R) = ₀ I 4 R The ratio of the magnetic fields is: B_P B_Q = ₀ I 8 R ₀ I 4 R = 4 8 = 1 2 Thus, the ratio is 1:2 . Answer: 1:2