A 1.8 kg copper rod rests on two horizontal rails (see figure below) 2.3 m apart

Question

A 1.8 kg copper rod rests on two horizontal rails (see figure below) 2.3 m apart and carries a current of 70 A from one rail to the other. The coefficient of static friction between rod and rails is 0.59. What is the magnitude and angle from the (counterclockwise from the x axis) of the smallest magnetic field (not necessarily vertical) that would cause the rod to slide? (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 will slide to the right.)

Explanation / Answer

friction force being applied on the rod=friction coefficient*normal force

=0.59*1.8*9.8=10.4076 N

hence the magnetic force on the rod has to be equal or greater in order for the rod to start moving.

let the rod slides to the +ve x direction.

hence friction will be along -ve x direction.

hence we need magnetic force to be along +ve x direction.

magnetic force is given by

F=I*(cross product of L and B)

where L=vector along the direction of current=+ve y direction

hence to get net force along +ve x direction, we need B to be along +ve z direction

i.e. out of the page(in first image)

hence, I*L*B=10.4076

==>70*2.3*B=10.4076

==>B=0.0646 T=64.6 mWb/m^2

hence the magnetic field required is 64.6 mWb/m^2 and along +ve z direction.

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