A rod of m = 135 kg rests on two parallel rails that are L = 025 m apart. The ro

Question

A rod of m = 135 kg rests on two parallel rails that are L = 025 m apart. The rod carries a current going between the rails (bottom to top in the figure, into the page) with a magnitude I = 42 A. A uniform magnetic field of magnitude B = 035 T pointing upward is applied to the region, as shown in the figure. The rod moves a distance d = 0.85 m along the rails in a given time period. Ignore the friction on the rails. Express the magnitude of the magnetic force F on the rod, in terms of the length L, current I, and magnetic field B. Calculate the numerical value of F, in newtons. Which direction does the rod move? Determine the work done by the magnetic force on the rod during its motion, in terms of the magnetic force F and the distance traveled d. Calculate the work done on the rod, in joules. Express the final speed of the rod, v, in terms of F, d, and m, if it started from rest. Calculate the numerical value of v, in meters per second. v =

Explanation / Answer

a)

magnetic force on the rod is given as

Fb = i B L Sin90

Fb = i B L

B)

Fb = i B L

inserting the values

Fb = 4.2 x 0.35 x 0.25

Fb = 0.3675 N

c)

using right hand rule , the magnetic force on rod acts in right direction , hence the rod moves towards right.

d)

work done is given as

W = Fb d Cos0

W = Fb d

e)

W = Fb d

inserting the values

W = 0.3675 x 0.85

W = 0.31 J

f)

Vi = initial velocity = 0

Vf = final velocity = v

m = mass of rod

using work-change in kinetic energy theorem

W = (0.5) m (Vf2 - Vi2)

F d = (0.5) m v2

hence , v = sqrt(2Fd/m)

g)

v = sqrt(2Fd/m) = sqrt(2 x 0.3675 x 0.85 /1.35) = 0.68 m/s

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