Question

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 4T and is directed out of the screen of the figure. The length of the wire is 5 cm and its mass mis 10 g. What current will produce a magnetic force on the wire such that it triples 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) Ww MWI 1. current-carrying wire L

Answer 1

Tension force is given by the weight mg of the wire. To triple the tension magnetic force must act downward.

Now, weight = mg = (10*10^-3*9.8) = 0.098 N

Magnetic force, F_m = ILB*sin90^o

= (I*5*10^-2*4*1)

= 0.2I

This force must act downward  to balance the tension of wire, which needs the current in the wire to flow from left to right, in accordance with right hand thumb rule. If mg were the only force downward, then, 2T = mg

When, tension in each lead becomes triple, net upward force = 3T+3T = 6T, means extra 4T force should be provided by magnetic force,

So, we must have, F_m = 2mg

Or, 0.2I = 2*0.098

Or, I = 0.98

So, answer is 0.98A, from left to right.