Question

A 2.4 kg copper rod rests on two horizontal rails (see figure below) 1.2 m apart and carries a current of 65 A from one rail to the other. The coefficient of static friction between rod and rails is 0.51. 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.)

A 2.4 kg copper rod rests on two horizontal rails (see figure below) 1.2 m apart and carries a current of 65 A from one rail to the other. The coefficient of static friction between rod and rails is 0.51. 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 .114 XT direction 62.978 o counterclockwise from the +x direction Rail Rod Rail hod Rail

Answer 1

The direction of magnetic field with respect to the vertical and the force due to it is shown in the following figure. Also the frictional force and the normal reaction and its weight are shown. The force diagram in the direction perpendicular to the page is also shown The Net force in the vertical direction is FNF3sin 0 mg = 0Fy+ Bilsin0- mg 0 FNmg -Bil sin 0 The net force in the horizontal direction is f Bil cos 0 uFy F3 cos 0 (mg-Bilsin) Bil cos 0+ uBilsin0 = umg Bil cos e иmg В - il (cos usin To find the minimum we must differentiate this and set this equal to zero я fo dB d иmg cos 0usin il (-sin0+ /ucos 0) 0 de il cos de sin 0) -1 0 tan T= tan0.51 e 27 Now the magnetic field is а. 0.51x2.4 x9.8 may В - 65x1.2cos 27° +0.51sin 27° B 0.137 T b. The angle is already calculated and it is 0 27