A 2.1 kg copper rod rests on two horizontal rails (see figure below) 1.7 m apart

Question

A 2.1 kg copper rod rests on two horizontal rails (see figure below) 1.7 m apart and carries a current of 55 A from one rail to the other. The coefficient of static friction between rod and rails is 0.63. What are the magnitude and angle (relative to the vertical) of the smallest magnetic field that puts the rod on the verge of sliding? (Based on the bottom picture, define to the right as the + x direction and up as the + y direction. Assume the current in the bottom picture is into the page and the magnetic force will cause the rod to slide to the right.) magnitude direction degree counterclockwise from the +x direction

Explanation / Answer

by FBD

N = mg-Fsintheta

in horizontal direction

Fcostheta = mu*N

F = mu*m*g/[costheta+mu*sintheta]

diffrentiating with respect to theta

dF/dtheta = mu*m*g(sintheta-mu*costheta)/(costheta+mu*sintheta)^2

is equal to zero when

sintheta = mu*costheta

tantheta = mu

theta = tan^-1(u) = tan^-1(0.63) = 32.21 degree

B = F/iL

B = mu*m*g/[costheta+mu*sintheta]*iL

B = 0.63 * 2.1 * 9.8 / ([ cos32.2 + 0.63 * sin32.2] * 55* 1.7)

B = 0.117 T = 0.12 T

theta = 32.2 degree

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