Practice question
Question
A solenoid with 1100 turns per meter carries \( 1.8 \, \text{A} \). What is the magnetic field inside?
(\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))
Explanation
**SI unit of magnetic field** is tesla (T), defined as force 1 N on 1 A·m wire perpendicular to field. Moving coil galvanometer uses torque τ = N I A B balanced by spring torque k φ, so deflection φ ∝ I, enabling current measurement, with radial field ensuring τ = N I A B always maximum. B = μ₀ n I . B = 4 π × 10⁻⁷ × 1100 × 1.8 = 7.92 π × 10⁻⁴ ≈ 2.49 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀
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