Question

Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is 6.03 × 10 − 5 N / m . Find the separation distance d of the wires expressed in millimeters.

Answer 1

d = 77.5mm

Step-by-step explanation:

The expression used for calculating the force between two parallel wires carrying currents separated by a distance is;

F12 =µµol1I2L/2πd where

I1 and I2 are the currents in both wires

L is the length of the wires

d is the separation distance

F12 is the magnitude of the force between them.

µo is the constant of proportionality

The equation can be rewritten as;

F12/L = µµol1I2/2πd

F12/L is the force per unit length acting on each wire= 6.03×10^-5 N/m

I1 = 3.35A

I2 = 6.99A

µ = 1

µo = 4π×10^-7N/A²

d =?

Substituting this datas in the equation above to get the separation distance d we have;

6.03×10^-5 =1× 4π×10^-7 × 3.35 × 6.99/2πd

6.03×10^-5 = 4×10^-7×23.42/2d

6.03×10^-5 = 9.37×10^-6/2d

2d = 9.37×10^-6/6.03×10^-5

2d = 1.55 × 10^-1

d = (1.55 × 10^-1)/2

d =0.155/2

d = 0.0775m

Since 1m = 1000mm

0.0775m = (0.0775×1000)mm

d = 77.5mm

Answer 2

244mm

Step-by-step explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m