# What is the magnitude of the net force on the first wire in (Figure 1)?

By Rebecca

What is the magnitude of the net force on the first wire in (Figure 1)?

You can see from the diagram that, wire 1 has an attracting force from wire 3 at 0.04 m distance and a repelling force from wire 2 at 0.02 m distance.

Expression of the force is –

F = µI1I2*L/(2πd)

When d = 0.02 m,

F = (4pi10^-710100.50) / (2pi*0.02) = 0.0005 N

And when d = 0.04 m,

F = (4pi10^-710100.50) / (2pi*0.04) = 0.00025 N

(a) So, the net force on wire 1 = 0.005 – 0.00025 = 0.00025 N repelling, or upward.

(c) By symmetry the net force on wire 3 is the same means, 0.00025 N, but downward direction.

(b) Wire 2 has equal repelling forces from wires 1 and 3 so net force = 0.

What is the magnitude of the net force on the first wire in (Figure 1)?

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