Force Between Parallel Conductors Problems and Solutions

 Problem#1

Figure 1 shows wire 1 in cross section; the wire is long and straight, carries a current of 4.00 mA out of the page, and is at distance d₁ = 2.40 cm from a surface.Wire 2, which is parallel to wire 1 and also long, is at horizontal distance d₂ = 5.00 cm from wire 1 and carries a current of 6.80 mA into the page. What is the x component of the magnetic force per unit length on wire 2 due to wire 1?

Fig.1

Answer:
We use
F_ba = i_bL x B_a

Force on wire b due to magnetic field produced by a

Fig.2

the magnetic field at point 2 due to wire 1 is

B₁ = μ₀i₁/2πR

With R = √(d₁² + d₂²), when

B₁ = μ₀i₁/{2π√(d₁² + d₂²)}

Therefore
F₂₁ = i₂LB₁
F₂₁/L = μ₀i₁i₂/{2π√(d₁² + d₂²)}

With (d₁² + d₂²) = [(2,40 cm)² + (5.00 cm)²] = 30.78 cm² = 3.078 x 10^─3 m²
And cos θ = d₂/√(d₁² + d₂²)

the x component of the magnetic force per unit length on wire 2 due to wire 1 is

(F₂₁/L)_X = μ₀i₁i₂ cos θ /{2π√(d₁² + d₂²)}
(F₂₁/L)_X = μ₀i₁i₂d₂/{2π(d₁² + d₂²)}
(F₂₁/L)_X = (4π x 10^-7 Tm/A)(4.0 x 10^─3 A)(6.8 x 10^─3 A)(0.05 m)/{2π x (3.078 x 10^─3 m²)}
(F₂₁/L)_X = 8.84 x 10^─11 N/m

Problem#2
In Fig. 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3. In unitvector notation, what is the net magnetic force per meter of wire length on wire 4?

Fig.3

Answer:

We use:
F₁₂ = μ₀Li₁i₂/2πr

 and superposition of forces:

F₄ = F₄₁ + F₄₂ + F₄₃

With θ = 45⁰, the situation is as shown below.

Fig.4

The components of F₄ are given by

F_4x = ─F₄₃ ─ F₄₂ cos θ  and F_4y = F₄₁ ─ F₄₂ sin θ

With F₄₃ = F₄₁ = μ₀i²L/2πa; F₄₂ = μ₀i²L/2√2πa; then

F_4x = ─[μ₀i²L/2πa] ─ [μ₀i²L/2√2πa] cos 45⁰
F_4x = ─3μ₀i²L/4πa

and
F_4y = [μ₀i²L/2πa] ─ [μ₀i²L/2√2πa] sin 45⁰
F_4y = μ₀i²L/4πa

Thus,
F₄ = {F_4x² + F_4y²}^1/2
F₄ = {(─3μ₀i²L/4πa)² + (μ₀i²L/4πa)²}^1/2
F₄ = (μ₀i²L√10)/4πa

F₄/L = (4π x 10^-7 Tm/A)(7.50 A)²/(4π x 0.135 m) = 1.32 x 10^─4 N/m

and F₄ makes an angle α with the positive x axis, where

α = tan^─1{F_4y/F_4x} = tan^─1{─1/3} = 162⁰

in unit vector notation, we have

F₄/L = (1.32 x 10^─4 N/m)(cos 162⁰i + sin 162⁰j)
F₄/L = (─1.25i + 0.42j) x 10^─4 N/m    

Problem#4
Figure 5 shows, in cross section, three currentcarrying wires that are long, straight, and parallel to one another. Wires 1 and 2 are fixed in place on an x axis, with separation d. Wire 1 has a current of 0.750 A, but the direction of the current is not given. Wire 3, with a current of 0.250 A out of the page, can be moved along the x axis to the right of wire 2. As wire 3 is moved, the magnitude of the net magnetic force F₂ on wire 2 due to the currents in wires 1 and 3 changes. The x component of that force is F_2x and the value per unit length of wire 2 is F_2x/L₂. Figure 5b gives F_2x/L₂ versus the position x of wire 3.The plot has an asymptote F_2x/L₂ = ─0.627 mN/m as x → ∞. The horizontal scale is set by x_s = 12.0 cm. What are the (a) size and (b) direction (into or out of the page) of the current in wire 2?

Fig.5

Answer:
At x = 4 cm, F_2x = 0, so i₁ is in the same direction as i₃. When i₃ approaches i₂ , F₂ is positive and dominated by the force from i₃ , so i₂ is out of the page.

Fig.6

At x = 4 cm:

F_2,net/L₂ = 0 = F₂₃/L₂ + (─F₂₁/L₂)

With, F₂₁/L₂ = μ₀i₁i₂/2πd and F₂₃/L₂ = μ₀i₂i₃/2πx, therefore:

μ₀i₁i₂/2πd = μ₀i₂i₃/2πx
d = i₁x/i₂ = (0.75 A/0.25 A) x 4 cm = 12.0 cm

When x →∞:

F_2,net/L₂ = F₂₁/L₂
μ₀i₁i₂/2πd = 0.627 x 10^─6 N/m
(4π x 10^-7 Tm/A)(0.75 A)(i₂)/(2π x 0.12 m) = 0.627 x 10^─6 N/m
i₂ ≈ 0.50 A

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