A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?

Respuesta :

Answer:

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction

Explanation:

Force per unit length between two current carrying wires is given by the formula

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

here we know that

[tex]F = 7.83 \times 10 N/m[/tex]

[tex]d = 7.0 cm = 0.07 m[/tex]

[tex]i_1 = 28 A[/tex]

now we will have

[tex]F = \frac{4\pi \times 10^{-7} (28.0)(i_2)}{2\pi (0.07)}[/tex]

[tex]7.83 \times 10 = \frac{2\times 10^{-7} (28 A)(i_2)}{0.07}[/tex]

[tex]i_2 = 978750 A[/tex]

Since the force between wires is attraction type of force so current must be flowing in upward direction