The inductance is given by L = Mu0 N^2 A/t, where Mu0 is the constant permeabili

Question

The inductance is given by L = Mu0 N^2 A/t, where Mu0 is the constant permeability of free space, N is the number of turns, t is length, and is the cross sectional area of the solenoid. Substituting A = pi r^2, where r is the radius, and the above expression for l into the equation for self-induced emf, we have the following. Epsilon_L = -L dI/dt = -(Mu0 N^2 pi r^2/l) dI/dt Remember: the current decreases uniformly over time so the value of dI/dt is negative. Solving for the radius, and substituting the given values, we have r = (-epsilon_L l/Mu0 N^2 pi(dI/dt))^1/2 = (-(175 time sign 10^-6v)(.139 m)/4pi time sign 10^-7 N/A^2)(435)^2 pi(-0.421 A/s))^1/2 = mm.

Explanation / Answer

r = {(-175 * 10-6) * 0.139 / [(4 * 10-7) * 4352 * * (-0.421)]}1/2 = 8.8 mm

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