Practice question
Question
A solenoid with 1400 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic
field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))
Explanation
**Torque on current loop** in magnetic field B is τ = N I A × B, magnitude τ = N I A B sinθ, N turns, I current (A), A area (m²) = l×b for rectangular, θ angle between normal to plane and B. Maximum when plane parallel to B (θ=90°), zero when perpendicular (θ=0°), magnetic moment m = N I A direction along normal via right-hand rule. Magnetic field B = μ₀ n I . B = 4 π × 10⁻⁷ × 1400 × 2.5 = 14 π × 10⁻⁴ ≈ 4.40 × 10⁻³ T . Using F = q v B sinθ, F
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