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Micheal Johnson

The figure is a cross section through three long wires with linear mass density 55.0 g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. (Figure 1)

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Answer:

The current that will allow the upper wire to float so as to form an equilateral triangle with the lower wires is 328.3 A.

Net force at the center of the equilateral triangle

The current that will allow the upper to float will form zero resultant force due to the current at the left and the current at the right.

Weight effect of the wires acting downwards

W =  mg

W = (σL)g

Resultant electric force of the two upward wires

F_1 = F_2 = \frac{\mu_o I^2 l}{2\pi r}

The total force due to angle formed by the top wire

F_t = F_1cos\theta + F_2 cos \theta=2Fcos\theta = \frac{2\mu_o I^2 l \times cos\theta}{2\pi r} \\\\F_t = \frac{\mu_o I^2 l \times cos\theta}{\pi r}

At equilibrium the upward force must equal downward force

\sigma lg=\frac{\mu_o I^2 l \times cos\theta}{\pi r}\\\\I^2 = \frac{\sigma lg \pi r}{\mu_0 l \times cos\theta} = \frac{\sigma g \pi r}{\mu_0 \times cos\theta} \\\\I^2 = \frac{(0.055)\times 9.8\pi \times 0.04 }{4\pi \times 10^{-7} \times cos60} \\\\I^2 = 107,800\\\\I = \sqrt{ 107,800} \\\\I = 328.3 \ A

Thus, the current that will allow the upper wire to float so as to form an equilateral triangle with the lower wires is 328.3 A.

Complete question is below:

The figure is a cross section through three long wires with linear mass density 55.0 g/m. They each carry equal currents in the directions shown. The lower two wires are 4.0 cm apart and are attached to a table. (Figure 1). What current I will allow the upper wire to "float" so as to form an equilateral triangle with the lower wires?

Learn more about magnetic force here:

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