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A wire with a weight per unit length of 0.071 N/m is suspended directly above a second wire. The top wire carries a current of 72.8 A, and the bottom wire carries a current of 72.7 A. The permeability of free space is 4π × 10−7 T · m/A . Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. Answer in units of mm.

Respuesta :

Answer:

d = 15 mm

Explanation:

Force of repulsion between two current carrying wire is given by

[tex]F = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now this force of repulsion is counterbalanced by the weight of the wire

so we have

[tex]mg = \frac{\mu_0 i_1 i_2 L}{2\pi d}[/tex]

now we have

[tex]d = \frac{\mu_0 i_1 i_2 L}{2\pi mg}[/tex]

so here we can say that

[tex]d = \frac{\mu_0 i_1 i_2}{2\pi (m/L)g}[/tex]

now plug in all values in it

[tex]d = \frac{4\pi \times 10^{-7} (72.8)(72.7)}{2\pi (0.071)}[/tex]

[tex]d = 0.015 m[/tex]

d = 15 mm

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