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Question

A solenoid has 600 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic
field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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Explanation

**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 × 600 × 3 = 7.2 π × 10⁻⁴ ≈ 2.26 × 10⁻³ T . Using F = q v B sinθ, F

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