A metal wire of mass m = 0.200 kg slides without friction on two horizontal rail
ID: 1460621 • Letter: A
Question
A metal wire of mass m = 0.200 kg slides without friction on two horizontal rails spaced a distance d = 0.08 m apart, as in the figure. The track lies in a vertical uniform magnetic field B = 1.20 T. There is a constant current i = 0.15 A through generator G, along one rail, across the wire, and back down the other rail. Find the speed and direction of the wire's motion as a function of time, assuming it to be stationary at t = 0. Evaluate for t = 0.4 s. Take positive to the right and negative to the left.
Explanation / Answer
Gravitational Potneital Energy U = mgh
as st max heght V = 0
Utotal = GPE = mgh = KE = 0.5 * 1 * 8.5 * 8.5 = 36.125 J
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Hmax = Hm = 36.125/(1 * 9.8)
Hm = 3.68 m from ground
height reached = 3.68 - 1.85 = 1.83 m
Magnetic force acting on the metal wire, F = i L B sin(90)
where i is current
L is length = d = 0.08 m
B is magnetic fiedl
so
F = 0.08*0.15 * 1.2 = 0.0144 N (towards left)
but from Force = ma
a = acceleration of metal wire, a = F/m = 0.0144/0.2 = 0.072m/s^2
speed at time t seconds,
v = u + a*t
= 0 + 0.071*t
V = 0.072*t
direction : towards left (-x axis)
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at t = 0.4 s
v = 0.072* 0.4
V = 0.0288 m/s