Respuesta :
Answer:
F_2 = 20 x 10⁻⁶ N
Explanation:
given,
current in right direction
I₁ = 10 A
I₂ = 5 A
I₃ = 8 A
Length of Wire 2 = 15 cm
distance between wire 1 and wire 2, r₁ = 5 cm
distance between wire 2 and 3 = r₂ = 17 - 5 = 12 cm
Force on wire 2 due to current in wire 1 and wire 2
[tex]F_2 = B_1I_2L_2 - B_3I_2L_2[/tex]
[tex]F_2 =(B_1 - B_3)I_2L_2[/tex]
[tex]F_2 =\dfrac{\mu_0}{4\pi}(\dfrac{2I_1}{r_1} -\dfrac{2I_3}{r_3} )I_2L_2[/tex]
μ₀ =absolute permeability of free space= 4π x 10⁻⁷ H/m
[tex]F_2 =\dfrac{4\pi \times 10^{-7}}{4\pi}(\dfrac{2\times 10}{0.05} -\dfrac{2\times 8}{0.12})\times 5\times 0.15[/tex]
F_2 = 20 x 10⁻⁶ N
