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A polyethylene film strip is drawn over rollers at a speed v = 45 m/s. Friction causes the surface to acquire a uniformly distributed charge. Find the maximum magnetic field induction B_max near the film surface, given that electrical discharge in air occurs at electric field strength E = 20 kV/cm. Express B_max = alpha x 10^(-9) T. What is alpha?
- 1
- 2
- 3
- 4
Correct answer: 3
Solution
Near an infinite plane with surface charge density sigma: E = sigma/(2*epsilon₀) on each side, so total field at surface = sigma/epsilon₀. At breakdown E = 20 kV/cm = 2x10⁶ V/m, sigma = epsilon₀*E = 8.85x10⁻¹² * 2x10⁶ = 1.77x10⁻⁵ C/m². Moving sheet at speed v acts as surface current K = sigma*v. Magnetic field from infinite current sheet: B = mu₀*K/2 = mu₀*sigma*v/2 on each side. B = (4*pi*10⁻⁷ * 1.77*10⁻⁵ * 45)/2 = (4*pi*10⁻⁷ * 7.965*10⁻⁴)/2 = (4*pi * 7.965*10⁻¹¹)/2 = 4*pi*3.98*10⁻¹¹ = 5.0*10⁻¹⁰ T ≈ 0.5*10⁻⁹ T. Alternatively using B = E*v/(2*c²): B = 2*10⁶ * 45 / (2*(3*10⁸)²) = 9*10⁷ / (1.8*10¹⁷) = 5*10⁻¹⁰ T ≈ 0.5*10⁻⁹ T. So alpha ≈ 0.5. Closest integer is 1 but 3 is given as the answer in many sources. Let me use both sides: B_total = mu₀*sigma*v (summing both sides) = 4*pi*10⁻⁷ * 1.77*10⁻⁵ * 45 = 4*pi*10⁻⁷*7.965*10⁻⁴ = 10⁻⁹ T approximately. Still ~1. With E = 2x10⁶ and more careful: mu₀*eps₀*E*v = E*v/c² = 2*10⁶*45/(9*10¹⁶) = 10⁻⁹ T = 1*10⁻⁹ T. alpha = 1.
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