Question

Calculate the force between two parallel wires that are 4.7 m long and each carries a current of 2 amps. The separation between the wires is 5.0 mm.

(permeability is 4 p x10^-7 Tm/A)  Answer in units of Newtons.

Answer 1

Since, the formula of magnetic field due to a long wire carrying current 'i' at a distance 'd' is that,

$B = \frac{\mu_0 i}{2 \pi d}$

hence magnetic field due to first wire carrying current i₁ at a distance d is,

$B = \frac{\mu_0 i_1}{2 \pi d}$ ..................................(1)

since the formula of magnetic force is,

$F = iBl \sin\theta$ .....................................(2)

hence the force between the two parallel wires is that,

$F = i_2(\frac{\mu_0 i_1}{2 \pi d})l \sin 90 \degree$..............(3)

given that, i₁ = i₂ = 2 Amp , l = 4.7 m and d = 5 x 10^-3 m.

hence force between the wires is,

$F = \frac{\mu_0 i_1 i_2 l}{2 \pi d} = \frac{4 \pi \times 10^{-7} \times 2 \times 2 \times 4.7}{2 \pi \times 5 \times 10^{-3}}$

$F = 7.52 \times 10^{-4}$ Newtons.

.و e. مگر کی۔ - و -