Question

D Question 6 5 pts A horizontal length of current-carrying wire is suspended from two identical flexible leads that are under tension due to the weight of the wire. The wire is oriented at right angles to a uniform magnetic field that has a magnitude of 4 T and is directed out of the screen of the figure. The length L of the wire is 5 cm and its mass mis 10 g. What current will produce a magnetic force on the wire that is just sufficient to remove the tension in the flexible leads, and which way is the current flowing through the wire? Current-carrying wire suspended from identical flexible leads Ceiling BO magnetic field (out of the screen) I, current-carrying wire L 0.18 A, current flows left to right 0.18 A, current flows right to left 0.45 A, current flows left to right 0.25 A. current flows right to left 0.25 A, current flows left to right

Answer 1

10x108kg x 9.8 m/s tension on each string is half of weight of wire 6 Force on the conduite F=Bilsino weight of wire, F=ing.. Foro remove tension Bilsino i= mg 2x Bisine 2x 4T* 5x10 masingo = 0.245 = 0.25 A Forie down out of plane soit right [fleming's left hand sule) 1+1+ Photo & 2 + 4 +3. 2+ ht? 9 / Rtotal = 12 3 R48R3 series R 34 = B 3 + Ry = 3+3=tr R34&R, & R, Parallel 4 - 1.32 9