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 4T 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 SBO magnetic field (out of the screen) 1, current-carrying wire L O 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

o L - BCee FB = il B so il B = mg (x 0.0544 = (10x163x0.8) 2 ( 0.2 SA from right hand thombrule righto to left) CS Scanned with Cam Scanner