Practice question
Question
A solenoid with 1300 turns per meter carries \( 1.2 \, \text{A} \). What is the magnetic field inside?
(\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))
Explanation
**Solenoid field** inside long solenoid is B = μ₀ n I, n = N/L turns per meter (m⁻¹), uniform and parallel to axis, outside negligible for long solenoid because fields from opposite sides cancel. For n = 1200 m⁻¹, I = 1 A, B = 4π×10⁻⁷×1200 = 1.51×10⁻³ T = 1.51 mT. B = μ₀ n I . B = 4 π × 10⁻⁷ × 1300 × 1.2 = 6.24 π × 10⁻⁴ ≈ 1.96 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =
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