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?

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

Answer 1

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The tension in the flessible leads can be removed with there is equal and opposite magnetic force is appled to guavitational force Famagueter = Fg = BIL Fo = 4x I x 5x102 m FB = 0.2 I L=5cm=5x102m FB Foravitational (mg) 10x10 3 x 9.8 mg I = 10x9-8810 3 = 0.490 A I N 0.2 I 0-2 I 0.490 FBD IFB wire I êx (fl) x ma a> I Aus was 80 0.490 A of current should flow in wire towards negatue a aris (left)