A metal wire of mass m = 0.200 kg slides without friction on two horizontal rail
ID: 1461572 • 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
the magnetic force is,
Fb = ILB = (0.15 A )( 0.08m )(1.20 T ) = 0.0144 N
The magnetic force is pointed to the left.
The acceleration interms of force is,
a = F/ m = 0.0144 N / 0.2 = 0.072 m/s2
the final speed is, Vf = at = 0.072*0.4 = 0.0288 m/s
This pointed to the left, so Vf = - 0.0288 m/s