A simple magnetic rail launcher consists of two metal rails separated by 5.00 cm
ID: 1325240 • Letter: A
Question
A simple magnetic rail launcher consists of two metal rails separated by 5.00 cm that are connected to a constant current source outputting 4,000 Amps. The constant current source ensures that the current through the rail remains at 4,000 A regardless of the voltages that arise due to the motion of the rail. {back emf} The rails are laid perpendicular to a uniform magnetic field. The magnitude of the field is 0.95 T. The object to be launched is a 0.055 kg piece of copper rod that electrically connects the two rails and slides without friction. The resistance of the rod is 0.0025 Ohms.
A simple magnetic rail launcher consists of two metal rails separated by 5.00 cm that are connected to a constant current source outputting 4,000 Amps. The constant current source ensures that the current through the rail remains at 4,000 A regardless of the voltages that arise due to the motion of the rail. {back emf} The rails are laid perpendicular to a uniform magnetic field. The magnitude of the field is 0.95 T. The object to be launched is a 0.055 kg piece of copper rod that electrically connects the two rails and slides without friction. The resistance of the rod is 0.0025 Ohms. A) ?Calculate the final velocity of the copper projectile if the metal rails are 10 meters long and tilted at an angle of 45 degrees. (ignore friction).Explanation / Answer
Note that the rail will thus carry the same current as the wires, I = 4000 A.
As F = BIL, then the force due to the B field is
F(B) = 0.95 T * 4000 A * 0.05 m
F(B) = 190 N, directed up the incline.
Now, note that there is a weight component down the incline, mgsin45:
F_gx = 0.055 kg * 9.8 m/s^2 * sin45
F_gx = 0.381 N
Thus, the net force uo the incline is
Fnet = 190 N - 0.381 N = 189.62 N
Thus, the net work done up the incline, done at a distance of 100 m, is
Fnet * d = 189.62 N * 10 m = 1896.2 J
Note that
Wnet = KEf - KEi
As KEi = 0, then
Wnet = 1896.2 J = 1/2mvf^2
Solving for vf,
vf = sqrt(2Wnet/m)
= sqrt(2*1896.2J/0.055kg)
= 260 m/s [ANSWER]
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