A metal wire of mass m = 0.300 kg slides without friction on two horizontal rail
ID: 2026802 • Letter: A
Question
A metal wire of mass m = 0.300 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 = 0.90 T. There is a constant current i = 0.30 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 = 0.6 s. Take positive to the right and negative to the left.
Please solve
Explanation / Answer
The force on the wire will come from the moving charged particles ( current ) experiencing a magnetic force from the magnetic field, B.
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 CONSTANT current, then we can ignore the effects of induction.
Therefore, Fb = ILB Fb = (.3 A )( .08m )( .9 T ) = .0216 N
The direction of the force is LEFT. using right hand rule vXB
Now, F = ma, so .0216 N = ( .3 kg )a
solving for a gives .072 m/s2
Now, finally, a fuction with respect to time. Speed would be Vf = vi + at.
Vf = 0 + ( .072 m/s2 )(t) at t = .6 sec, so plug it into this equation to get .0432 m/s
The direction is left, so it's negative, so the answer is -.0432 m/s