Question

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 40.0 g and electrical resistance 0.200 Ωrests on parallel horizontal rails that have negligible electric resistance. The rails are a distance L = 10.0 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 = 7.84×10⁻² T . Assume that the rod has a slightly flattened bottom so that it slides instead of rolling. Use 9.80 m/s2 for the magnitude of the acceleration due to gravity.

Find μs, the coefficient of static friction between the rod and the rails.

Answer 1

Solution: From the question we have

A rail gun uses electromagnetic forces to accelerate a projectile to very high velocities

A metal rod of mass 40.0 g and electrical resistance 0.200 Ωrests on parallel horizontal rails that have negligible electric resistance.

The rails are a distance L = 10.0 cm apart.

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 = 7.84×10−2 T

To find:Find μs, the coefficient of static friction between the rod and the rails.

Current I in the rod = V/R = 5/0.2 = 25 A
Length through which B crosses = 0.08 m
Magnetic field B = 7.35e-2 (No unit is given)

Force acting on the rod = ILB = 25*0.1*7.84×10−2 T = 0.196 N (v=IR=I=V/R=5/0.2=25 A

Frictional force = μmg
Net force = 0.196 - μmg = 0 when motion just begins.

μ = 0.196 / (0.04*9.8) = 0.5 ANSWER