A long solid cylindrical conductor of radius R carries a steady current uniformly distributed over its cross-section. The magnitude of the magnetic field produced by this current is exactly half of its maximum possible value at two different radial distances from the axis: r₁ (inside the conductor) and r₂ (outside the conductor). The ratio r₂ r₁ is equal to:

Step-by-step solution

The maximum magnetic field for a long solid cylindrical conductor carrying a uniform current occurs at its surface ( r = R ) and is given by: B_ max = ₀ I 2 R Inside the conductor ( r B_ in = ₀ I r 2 R^2 Given that B(r₁) = B_ max 2 for r₁ ₀ I r₁ 2 R^2 = 1 2 ( ₀ I 2 R ) r₁ = R 2 Outside the conductor ( r > R ), the magnetic field is: B_ out = ₀ I 2 r Given that B(r₂) = B_ max 2 for r₂ > R : ₀ I 2 r₂ = 1 2 ( ₀ I 2 R ) r₂ = 2R The required ratio is: r₂ r₁ = 2R R/2 = 4 Answer: 4