A 1.10 kg copper rod rests on two horizontal rails 0.64 m apart and carries a cu
ID: 1604008 • Letter: A
Question
A 1.10 kg copper rod rests on two horizontal rails 0.64 m apart and carries a current of 50 A from one rail to the other. The coefficient of static friction between rod and rails is 0.54. What is the smallest magnetic field (not necessarily vertical) that would cause the rod to slide? What is the angle of B from the vertical? (deg) (in deg) 2.13 times 10^1 2.84 times 10^1 3.77 times 10^1 5.02 times 10^1 6.67 times 10^1 8.88 times 10^1 118 times 10^2 1.57 times 10^2 2.09 times 10^2 2.78 times 10^2Explanation / Answer
We know that
N=mg-Fsin and Fcos=N just before slipping
Solving for F yields
F=mg/(cos + sin)
Differentiating with respect to to above equation
dF/d = mg/(sin - cos)/( cos + sin)2 ,which is zero when sin = cos
or when tan==0.5 ==>=0.4636
F=(0.54x1.1x9.81)/(0.8944+(0.54x0.4472)
=5.82714/1.136
=5.13N
Since the current is out of phase ,for the magnetic field B in the plane of the phase at right angles to the desired force
Making an angle /2 + =2.304 with the horizontal x direction
F=iLBs
L=0.64m
Bs=F/iL
=5.13/(50x0.64)
=0.16T
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