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Question

A solenoid has 800 turns per meter and carries a current of \( 2.5 \, \text{A} \). What is the magnetic
field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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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. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 800 × 2.5 = 8 π × 10⁻⁴ ≈ 2.51 × 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

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