Under the influence of gravity, a rectangular wire loop of mass m, wifth w, leng
ID: 1439023 • Letter: U
Question
Under the influence of gravity, a rectangular wire loop of mass m, wifth w, length l, and resistance R falls out of a uniform magnetic field B. The loop is released from rest at the instant its lower edge is just barely inside the magnetic field out of the field. The loop is eciting the magnetic field at speed v(t) > 0. Velocity is downward in the same direction as acceleration. obtain an expression for the speeed of the loop as a function of time, for the time intercal from the instant the loop is released to the instant just before its top edge exits the magnetic field. what happens after the top edge of the loop las left the field.
Explanation / Answer
As long as the flux through the loop is changing, there is an upward magnetic force on the induced current in the bottom wire, which may cancel the downward force of gravity on the loop.
The flux through the loop is proportional to the vertical distance it falls into the field region
= BA = Bwy,
so Faraday’s and Ohm’s laws give a magnetic force proportional to the vertical speed,
F = IwB,
mg = d/dt*wB/R = vw^2B^2/R
v = mgR/w^2B^2
After crossing the field, the loop will go down due to gravity.
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