A metal rod of length 21.8 cm and mass 109 g is free to slide over two parallel,
ID: 1399389 • Letter: A
Question
A metal rod of length 21.8 cm and mass 109 g is free to slide over two parallel, horizontal metallic rails connected to a 10.0 V battery (see figure below). The battery has internal resistance of 0.100 , and the rod and rails have negligible resistance. A 1.94 T magnetic field is directed perpendicular to the plane of the circuit
(a) Find the magnetic force on the rod.
(b) Find the rod's speed after it has moved 28.2 cm, assuming negligible friction.
Explanation / Answer
a) The current produced is i = E / Rt .. .. 10.0V / 0.10
which gives the current, i = 100 A
Thus the Magnetic force produced is F = BiL
F = (1.94))(100A)(0.218m)
which gives force F = 42.3 N
Force direction can be obtained by applying Fleming's left-hand rule . By using the rule we can find the direction.
Field direction is into page (a convention indicated by the + symbols on diagram)
Force direction is along the rails which is away from battery.
b) Next we need to find the Rod acceleration . The rod accelereation is given as a = F/m
a= 42.3 N / 0.109kg
a = 388 m/s2
Initial velocity u is given as,u = 0
d = 0.282 m
v2 = u2 + 2ad
v2= 0 + (2 x 388m/s2 x 0.282m)
v2 = 218.83
which implies v = 14.79 m/s
So rods speed is 14.8 m/s.