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Two long parallel wires are separated by 15 cm. One of the wires carries a current of 34 A and the other carries a current of 69 A. The permeabilty of free space is 1.257 × 10−6 N · m/A. Determine the magnitude of the magnetic force on a 5.9 m length of the wire carrying the greater current. Answer in units of mN

Respuesta :

Answer:

F = 0.018 N

Explanation:

Magnetic force between two parallel current carrying wires is given by

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

here we know that

[tex]i_1 = 34 A[/tex]

[tex]i_2 = 69 A[/tex]

d = 15 cm

L = 5.9 m

now from above formula we can say

[tex]F = \frac{(4\pi \times 10^{-7})(34 A)(69 A)5.9}{2\pi (0.15)}[/tex]

now the force between two wires is given as

[tex]F = 0.018 N[/tex]

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