A metal wire of mass m = 0.500 kg slides without friction on two horizontal rail
ID: 2259871 • Letter: A
Question
A metal wire of mass m = 0.500 kg slides without friction on two horizontal rails spaced a distance d = 0.32 m apart, as in the figure. The track lies in a vertical uniform magnetic field B = 1.50 T. There is a constant current i = 0.20 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 = 0.5 s. Take positive to the right and negative to the left.
Explanation / Answer
The formula to solve for this is Fb = ILB
The motion of the wire will cause a change in flux, which will induce a current and fight the motion...however, since the generator allows for a constnt current, then we can ignore the effects of induction.
Therefore, Fb = ILB ? Fb = (0.2 A )( 0.32m )( 1.5 T ) = 0.096 N
The direction of the force is LEFT. using right hand rule VXB
Now, F = ma, so 0.096 N = ( 0.5 kg )a
solving for a gives a=0.192 m/s^2
Now, finally, a fuction with respect to time. Speed would be Vf = vi + at.
Vf = 0 + ( 0.192 m/s2 )(t) at t = 0.5 sec, so plug it into this equation to get 0.096 m/s
The direction is left, so it's negative, so the answer is -0.096 m/s