Question

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 m is 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?

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

Current-carrying wire suspended from identical flexible leads SBO magnetic field (out of the screen) Ceiling 1, current-carrying wire L

Answer 1

BO FB TA +T * 2 Tz mg INN IN T= mg 2 Given, B = 4T, mg 1=5cm = mzlog zloxio I = ? 125x102m -3 kg But, FB = T = mg/ ILB = mg I = mg 2LB 16x1x 9.8 I 2xsxxotx 4 9.8 40 = 0.245 A I = 6.245A correct choice, 0 0.25 A, Current flows right to left.