Inductance of a coil with 10⁴ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω. The energy density in the inductor when the current reaches (1/e) of its maximum value is απ × (1/e²) J/m³. The value of α i $s ___.$
(μ₀ = 4π × 10⁻⁷ Tm/A)

Question

Inductance of a coil with 10⁴ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω. The energy density in the inductor when the current reaches (1/e) of its maximum value is απ × (1/e²) J/m³. The value of α i $s ___.$

(μ₀ = 4π × 10⁻⁷ Tm/A)

G1 Physics · Accepted Answer

The maximum steady current supplied by the 10 V source through 10 Ω is I(max) = 10/10 = 1 A. At the stated instant, I = 1/e A. The magnetic-energy density in a solenoidal magnetic field is u = B²/(2μ₀). Using B = μ₀nI, we get u = μ₀n²I²/2. The supplied source solution evaluates the turn density as n = 10⁴ m⁻¹ and therefore uses n² = 10⁸ m⁻². Following the source calculation, u = [4π × 10⁻⁷ × 10⁸ × (1/e²)]/2 = 20π/e² J/m³. Thus, α = 20. Source-data note: the question text states 10⁴ turns, while…

Inductance of a coil with 10⁴ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of 10 Ω. The energy density in the inductor when the current reaches (1/e) of its maximum value is απ × (1/e²) J/m³. The value of α i s ___. (μ₀ = 4π × 10⁻⁷ Tm/A)