A rod of m = 0.55 kg rests on two parallel rails that are L = 0.35 m apart. The

Question

A rod of m = 0.55 kg rests on two parallel rails that are L = 0.35 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude 1 = 3.4 A. A uniform magnetic field of magnitude B = 0.85 T pointing upward is applied to the region, as shown in the graph. The rod moves a distance d = 1.45 m. Ignore the friction on the rails. Calculate the final speed of the rod if it started from rest, in m/s. Assume the current through the rod is constant at all times. v =_______

Explanation / Answer

I = 3.4 A
B = 0.85 T
m = 0.55 Kg
L = 0.35 m
d = 1.45 m

Force on the rod in magnetic Field, F = B*I*L
We know,
F = m*a
m*a = B*I*L
Substituing values,
0.55*a = 0.85*3.4*0.35
a = 1.84 m/s^2

Now,
initial velocity, u = 0
Vf^2 = u^2 + 2*a*d
Vf^2 = 0 + 2*1.84*1.45
Vf = 2.31 m/s
Final speed of the rod, Vf = 2.31 m/s

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