Force Between Parallel Conductors Problems and Solutions 2

 Problem#1

In Fig. 1, five long parallel wires in an xy plane are separated by distance d = 8.00 cm, have lengths of 10.0 m, and carry identical currents of 3.00 A out of the page. Each wire experiences a magnetic force due to the currents in the other wires. In unit-vector notation, what is the net magnetic force on (a) wire 1, (b) wire 2, (c) wire 3, (d) wire 4, and (e) wire 5?

Fig.1

Answer:

We use
F_ab = μ₀i_ai_b/2πr_ab

Two wires with parallel current attract each other. Here i₁ = i₂ = i₃ = i₄ = i = 3 A

(a) The magnetic force on wire 1 is
F₁ =  [μ₀i²L/2π][1/d + 1/2d + 1/3d + 1/4d] = 25μ₀i²L/24πd
    =  25 x (4π x 10^-7 Tm/A)(3.00 A)²(10.0 m)/(24π x 0.08 m)
F₁ = 4.69 x 10^-4 N to the top

(b) similarly, for wire 2. We have
F₁ =  [μ₀i²L/2π][1/2d + 1/3d] = 5μ₀i²L/12πd
    =  5 x (4π x 10^-7 Tm/A)(3.00 A)²(10.0 m)/(12π x 0.08 m)
F₂ = 1.88 x 10^-4 N to the top right

(c) F₃ = 0 (because of symetry)

(d) F₄ = –F₂ = –1.88 x 10^-4 N downward

(e) F₅ = –F₁ = –4.69 x 10^-4 N downward

Problem#2
In Fig. 1, five long parallel wires in an xy plane are separated by distance d = 50.0 cm. The currents into the page are i₁ = 2.00 A, i₃ = 0.250 A, i₄ = 4.00 A, and i₅ = 2.00 A; the current out of the page is i₂ = 4.00 A.What is the magnitude of the net force per unit length acting on wire 3 due to the currents in the other wires?

Answer:
The first thing you should do is carefully label all the information given to you on your diagram as shown in the modified diagram with red markings (this is what you would pencil in). The circle with a dot represents a field coming out of the page. The circle with an ”X” represents a field going into the page. This convention arises because it is thought that the circle with a dot in the middle looks like an arrow head approaching your face; the circle with the ”X” looks like the vanes of an arrow - what you’d see of an arrow moving away from you. Anyway, the job is to calculate the force on a wire. Our knee-jerk reaction is to recall that a current carrying wire exposed to an extermal B field is

F = i(L × B)

where L is the current direction which is situated in some external B . In our case, we want to look at the force on wire 3, which has a current of 0.25A

Fig.2

Figure 2: you should draw these currents and symbols on your paper. into the page. The force on i3 will result from the total B that wire 3 experiences due to the B fields produced by all the other wires. The recipe is:

• We calculate the B field B₁ due to i₁ at the position of wire 3.
• We calculate the B field B₂ due to i₂ at the position of wire 3.
• We calculate the B field B₄ due to i₄ at the position of wire 3.
• We calculate the B field B₅ due to i₅ at the position of wire 3.
• We sum to get effective field at wire 3: B_eff = B₁ + B₂ + B₄ + B₅
• We use B_eff in the equation for the force on wire 3: F_on3 = i₃(L₃ × B_eff)

Starting with B₁: The B field due to i₁ at the position of wire 3 is found by Ampere’s law. We know this result: B due to a current carrying wire at a distance R from the wire has the magnitude

B = μ₀i₁/2πR

and a direction given by the right hand rule: point your right thumb in the direction of the current, and curl your fingers around to your palm: the field that current produces wraps around the wire in the direction of the curly fingers. Anyway, getting the field due to wire 1 at the position of wire 3 requires we use the distance R = 2d

B = μ₀i₁/(2π x 2d)

plugging in the numbers, and noting since i₁ is into the page, B will point in the −z direction (down) at wire 3:

B = (4π x 10^-7 Tm/A)(2.0 A)/(2π x 2 x 0.5 m) = 4.0 x 10^-7 T (down)

While the field due to i₁ points down at wire 3, a quick examination with the right hand rule reveals that the field from i₁ points down at wire3, but the fields due to i₂, i₄, and i₅ all point up. So B₁ will subtract and the others will add to the effective B_eff . Now we are ready to truly grind this out. The distances of each from wire 3 are

d₁ = 2d = 1.0 m; d₂ = d = 0.5 m; d₄ = d = 0.5 m and d­₅ = 2d = 1.0 m

The current magnitudes,
i₁ = 2.0 A; i₂ = 4.0 A; i₄ = 4.0 A and i­₅ = 2.0 A

So recalling only B₁ points down, the B’s are
B₁ = –μ₀i₁/2πd₁ = (4π x 10^-7 Tm/A)(2.0 A)/(2π x 1.0 m) = 4.0 x 10^-7 T
B₂ = –μ₀i₂/2πd₂ = (4π x 10^-7 Tm/A)(4.0 A)/(2π x 0.5 m) = 1.6 x 10^-6 T
B₄ = –μ₀i₄/2πd₄ = (4π x 10^-7 Tm/A)(4.0 A)/(2π x 0.5 m) = 1.6 x 10^-6 T
B₅ = –μ₀i₅/2πd₅ = (4π x 10^-7 Tm/A)(2.0 A)/(2π x 1.0 m) = 4.0 x 10^-7 T

Therefore the total effective B_eff at the position of wire 3 is
B_eff = B₁ + B₂ + B₃ + B₄ = 3.2 x 10^-6 T up

and so the magnitude of the force on wire 3 is

F_on3 = i₃(L₃ × B_eff)

We don’t have L which is why we are asked the force per unit length:

F_on3/L₃ = i₃B_eff = 0.25 A x 3.2 x 10^-6 T = 8.0 x 10^-7 N

Problem#3
In Fig.3 a long straight wires carries a current i₁ = 30.0 A and a rectangular loop carries current i₂ = 20.0 A.Take the dimensions to be a = 1.00 cm, b = 8.00 cm, and L = 30.0 cm. In unit vector notation, what is the net force on the loop due to i₁?

Fig.3

Answer:
By symmetry, we note that the magnetic forces on the right and left segments of the rectangle cancel. The net force on the vertical segments of the rectangle is

F = F₁ + F₂

With F₁ = μ₀i₁i₂L/[2π(a + b)] and F₂ = μ₀i₁i₂L/2πa, then

Fig.4

F = ─μ₀i₁i₂L/[2π(a + b)] + μ₀i₁i₂L/2πa
F = [μ₀i₁i₂L/2π][1/a ─ 1/(a + b)]
   = [(4π x 10^-7 Tm/A)(30.0 A)(20.0 A)(0.30 m)/2π][1/0.01 m ─ 1/0.09 m]
F = 3.2 x 10^─3 N

In unit vector notation, the net force on the loop due to i₁ is
F = (3.2 x 10^─3 N)j = 3.2 mNj

Problem#8
Three long wires are parallel to a z axis, and each carries a current of 10 A in the positive z  direction.Their points of intersection with the xy plane form an equilateral triangle with sides of 50 cm, as shown in Fig. 5. A fourth wire (wire b) passes through the midpoint of the base of the triangle and is parallel to the other three wires. If the net magnetic force on wire a is zero, what are the (a) size and (b) direction (z or z) of the current in wire b?

Fig.5

Answer:
Forces acting on wire 1 due to wires 2, 3 and 4 are shown in the figure.

Fig.6

By symmetry: F₁₂ = F₁₃

the resultant force on wire 1 due to wire 2 and wire 3 is

F₁₂₃ = F₁₂ + F₁₃

With F₁₂ = F₁₃ = 𝜇₀𝑖₁𝑖₂/2𝜋d = 𝜇₀𝑖²/2𝜋d, then

F₁₂₃ = 2F₁₂ cos 30⁰
F₁₂₃ = 𝜇₀𝑖² cos 30⁰/2𝜋d (downward)

so that the resultant magnetic force acting on wire 1 is zero, then the magnetic force in wire 1 due to wire 4 must be directed upwards and the amount is equal to F₁₂₃. Therefore:

F₁₂₃ = F₁₄
𝜇₀𝑖² cos 30⁰/2𝜋d = 𝜇₀𝑖i₄/2𝜋r_ab
𝑖cos 30⁰/d = i₄/r_ab
i₄ = 𝑖r_ab cos 30⁰/d
because r_ab = [(d)² ─ (d/2)²]^1/2 = ½d√3, then
i₄ = ½√3 x 𝑖 x cos 30⁰ = ½√3 x 10 A x ½√3 = 15 A  

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