Question

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a downward current of 4.00 A . A third wire, wire 3 , is to be hung vertically and located such that when it carries a certain current, each wire experiences no net force. (c) the magnitude and direction of the current in wire 3 .

Answer 1

The magnitude of the current in wire 3  is 2.4 A and in a direction pointing in the downward direction.

• The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$  wires at distance ' r ' is given by  $f=\frac{u_0I_1I_2}{2\pi r}$   ....(1) .

• If the current is flowing in both wires in the same direction, and  the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$

I₃ = 2.4 A and the current is pointing in the downward direction

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