Class Management I Help hw 8 (Magnetic Force II) Begin Date: 11/8/2017 12:01:00
ID: 1783041 • Letter: C
Question
Class Management I Help hw 8 (Magnetic Force II) Begin Date: 11/8/2017 12:01:00 AM -- Due Date: 11/16/2017 11:59:00 AM End Date: 11/17/2017 11:59:00 PM (14%) Problem 7: A rod of m = 2.6 kg rests on two parallel rails that are L = 0.69 m apart. The rod carries a current going between the rails (bottom to top in the figure) with a magnitude I 1.2 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= 0.55 m. Ignore the friction on the rails ©theexpertta.com Calculate the final speed, in meters per second, of the rod if it started from rest, assuming there is no friction in the contact between it and rails. Assume the current through the rod is constant at all times Grade Summary 0% 100% Potential sin) cotan atan) acotan) sinh) cosh cosO tan() | Submissions Attempts remaining: 20 (4% per attempt) detailed view tanh0 cotanh0) Degrees Radians 0 END BA Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 3 Feedback: 0% deduction per feedback. All content © 2017 Expert TA, ILCExplanation / Answer
magnetic force experienced by rod Fb = I*L*B*sintheta
I = current inthe rod
L = length of the rod
B = magnetic field = 0.85 T
theta = anlge betwen L and B = 90
Fb = 1.2*0.69*0.85*sin90 = 0.7038 N
work done by force W = Fb*d
from work energy relation work = change in KE
W = (1/2)*m*(vf^2 - vi^2)
vi = initial speed = 0
0.7038*0.55 = (1/2)*2.6*(vf^2-0)
vf = sqrt(2*0.7038*0.55/2.6)
speed v = 0.545 m/s <<<<-------ANSWER