Question

A 0.300-kg metal rod carrying a current of 12.0 A glides on two horizontal rails 0.590 m apart and 2.0 m long. (a) If the coefficient of kinetic friction between the rod and rails is 0.160, what vertical magnetic field is required to keep the rod moving at a constant speed? (b) If the friction between the rod and rail is reduced zero, the rod will accelerate. If the rod starts from rest at the one end of the rails, what is the speed of the rod at the other end of the rails for this frictionless situation? Use the same field value you calculated in part (a).

Answer 1

For moving toward right force on ban = BIL Ton B НЕ B=magnetic field. 1 = current flowing through the bar = 12 A L = .590 meter. here force i towards right! So according right hand rule magnetic field is into the page frictional force is opposite to the direction of force. fk = lx R = dia kinetic friction Coefficient m= mass of Ma mga

fk - (0.160) (-300) (10) N [g-lo m/s .48 N. Force on unrrent carrying bar F = BIL =BX 12 X 550 For constant speed I Frer = 0 F = fk BX12X. 590 .48 Ba 48 12x. 590 6.78 Tesla -) An 6.78 Tesla, a magnetic field value = then If friction in hero, acceleration of bon BIL - (6-78) (12) (:590) m m • 300 160 m/s2 using the formula, ler-u2+2 as u=initial speed=omls. a = 160 m/s2 5- distanu length of rail = 2 m. 2 => 297 = 0 +2(160) (2) 1 = V2x16082 > 25.29 m/s. Am Speed of bar at other end = 25.29 m/s.