In the figure, a metal wire of mass m = 24.1 mg can slide with negligible friction

In the figure, a metal wire of mass m = 24.1 mg can slide with negligible friction on two horizontal parallel rails separated by distance d = 2.56 cm. The track lies in a vertical uniform magnetic field of magnitude 56.3 mT. At time t = 0, device G is connected to the rails, producing a constant current i = 9.13 mA in the wire and rails (even as the wire moves). At t = 61.1 ms, what are the wire’s (a) speed and (b) direction of motion (left or right)?

[stevenmay]

shouldn't mg be 24.1*10^-3 since it is milligrams not micro?

[BramhamBramham]

magnetic field B = 66.0 mT,
mass of metal wire m = 25.5 mg
distance d = 1.31 cm
current i = 7.04 mA
time t = 63.1 ms

we know that the magnetic force on a current carrying wire inside a magnetic field B is

F = B*i*l sin theta
here theta is 90 degrees, that is angle between the current and magnetic field

from Newton's second law F = ma

m*a = B*i*l sin theta

a = B*i*l sin theta /m


l is replaced by d

a = B*i*d sin theta /m


substituting the values
a = /(25.5*10^-3)


a = 0.0002387 m/s^2

from equations of motions

v = u+at
here u = 0 m/s

v = 0+a*t

v = 0+0.0002387*63.1*10^-3 m/s = 1.50618*10^-5 m/s

[jaketoczek]

Why can you ignore that B and L are vector quantities? Do you not have to do B cross L?

[BramhamBramham]

mass was substituted in micro grams it should be 24.1*10^-3 kg instead of 24.1*10^-6 kg