HW 12 Rail Gun 1 of 55 Constants A rail gun uses electromagnetic forces to accel
ID: 2032661 • Letter: H
Question
HW 12
Rail Gun
1 of 55
Constants
A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass 50.0 g and electrical resistance 0.300 ? rests on parallel horizontal rails that have negligible electric resistance. The rails are a distance L = 5.00 cm apart. (Figure 1)The rails are also connected to a voltage source providing a voltage of V = 5.00 V .
The rod is placed in a vertical magnetic field. The rod begins to slide when the field reaches the value B = 0.294 T . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use 9.80 m/s2for the magnitude of the acceleration due to gravity.
Constants
A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities. The basic mechanism of acceleration is relatively simple and can be illustrated in the following example. A metal rod of mass 50.0 g and electrical resistance 0.300 ? rests on parallel horizontal rails that have negligible electric resistance. The rails are a distance L = 5.00 cm apart. (Figure 1)The rails are also connected to a voltage source providing a voltage of V = 5.00 V .
The rod is placed in a vertical magnetic field. The rod begins to slide when the field reaches the value B = 0.294 T . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use 9.80 m/s2for the magnitude of the acceleration due to gravity.
Explanation / Answer
Using Force Balance:
Fnet = Fm - Ff = 0
Fm = Ff
Fm = magnetic force = i*L*B
from ohm's law:
i = V/R
Fm = V*L*B/R
Ff = frictional force = uk*m*g
uk*m*g = V*L*B/R
uk = V*L*B/(m*g*R)
Using given values:
uk = 5*5*10^-2*0.294/(50*10^-3*9.81*0.300)
uk = 0.499 = 0.5
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