Practice question

**Ampere's circuital law** ∮ B·dl = μ₀ I_enc explains solenoid field, choosing rectangular Amperian loop with one side inside, one outside where B≈0, giving B L = μ₀ n L I ⇒ B = μ₀ n I. Toroid B = μ₀ N I/(2π r) inside, zero outside, field confined. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 70 × 0.8/2 × 0.07) = (22.4 π × 10⁻⁶/0.14) = 1.6 π × 10⁻⁴ ≈ 5.03 × 10⁻⁴ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B