Skip to content

Question

A solenoid’s current is switched off suddenly. What causes sensitive instruments nearby to be at risk
of damage?

Options

Choose one · Correct answer highlighted

Explanation

**Energy stored in inductor** U =½ L I² (J), L inductance (H), I current (A), resides in magnetic field, energy density u = B²/(2μ₀) (J/m³), B field (T), μ₀=4π×10⁻⁷ H/m. For solenoid B=μ₀ n I, volume V= A l, U = (B²/2μ₀) V =½ (μ₀ n² A l) I² =½ L I², consistent. The sudden drop in current induces a high emf due to self-inductance, which can generate significant currents or voltages in nearby circuits, risking damage. Using Φ = B A cosθ, e = -N dΦ/dt = -N A dB/dt = B l v = N B A ω sinωt, L = μ₀ N²A/l,

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.