handin05 - Physics 212 Classical and Modern Physics Spring...

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Physics 212 Classical and Modern Physics Spring 2008 Solutions for Hand-In Set #5 Problem A28 — Falling Loops As the loop falls out of the magnetic field region, the flux through the loop decreases, and so an emf drives a current through the loop. We can write the flux as Φ B = NBA cos θ = B 0 wx, where N = 1, cos θ = 1, and x is the length of the loop still in the magnetic field, so that A = wx . The emf is then just the time derivative of the flux, E = d Φ B dt = B 0 wv, since dx dt = v , the speed at which the loop falls. The magnitude of the induced current is I = E R = B 0 wv R . Two forces act on the loop: the force of gravity acts down, and there is an upward directed magnetic force acting on the counterclockwise current. Newton’s second law gives F net = ma, or F mag - F grav = ma. The magnitude of the magnetic force is given by F mag = IwB = ± B 0 wv R ² wB 0 = B 2 0 w 2 v R which grows linearly with speed v . After the loop is dropped, the speed (and thus the magnetic force) grows until the magnetic force and the gravitational force are
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handin05 - Physics 212 Classical and Modern Physics Spring...

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