Question

Wires 1, 2, and 3 each have current moving through them to the right. I1 = 10 A, I2 = 5 A, and I3 = 8 A. Wire 2 is 15 cm long and 5 cm below wire 1. Wire 3 is 17 cm below wire 1. What is the force on wire 2?

Answer 1

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

$F_2 = B_1I_2L_2 - B_3I_2L_2$

$F_2 =(B_1 - B_3)I_2L_2$

$F_2 =\dfrac{\mu_0}{4\pi}(\dfrac{2I_1}{r_1} -\dfrac{2I_3}{r_3} )I_2L_2$

μ₀ =absolute permeability of free space=  4π x 10⁻⁷ H/m

$F_2 =\dfrac{4\pi \times 10^{-7}}{4\pi}(\dfrac{2\times 10}{0.05} -\dfrac{2\times 8}{0.12})\times 5\times 0.15$

   F_2 = 20 x 10⁻⁶ N