In the figure a metal wire of mass m 249 mg can slide with

In the figure, a metal wire of mass m = 24.9 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 1.44 cm. The track lies in a vertical uniform magnetic field of magnitude 56.7 mT. At time t = 0 s, device G is connected to the rails, producing a constant current i = 9.36 mA in the wire and rails (even as the wire moves). At t 59.5 ms, what are the wire\'s speed and direction of motion? Number Units the tolerance is +/-2%

Solution

The force is F=iL X B, they are all perpendicular so it is iLB. i=0.00936 A, L=0.0144 m, B=0.0567 T (Tesla).
F = 0.00936 * 0.0144 * 0.0567 [N] (force in Newtons)

=7.64 * 10^-6 N

Now: F=ma (mass times acceleration)
m=0.0249 kg
so a=F/m (we have already got the force so can calculate this)
the speed v = a*t (acceleration times time in seconds)
t=0.0595

so V = (F/m)*t = (7.64 * 10^-6 /.0249) * .0595 = 15.40 * 10^-6 m/s.

The direction would be left.