please answer :

Question

Two parallel wires, 4 cm apart, carry currents of 2 A and 4 A respectively, in the same

direction. The force per unit length in N/m of one wire on the other is:

Answer 1

Dorsa,

The equation you need to solve this question is

(force per unit length) = (vacuum permeability constant)*(current 1)*(current 2) / 2π(distance between wires)

which is symbolically written as

F/ΔL = (μ₀ I₁ I₂ )/(2π r)

If you search for "vacuum permeability constant" you'll find it μ₀ = 1.26×10⁻⁶ H/m (units are Henry/meter). Also 4 cm = 0.04 m.

Therefore,

force per unit length = (1.26×10⁻⁶ H/m)(2 A)(4 A) / (2π * 0.04 m)

force per unit length = (1.01×10⁻⁵ HA²/m) / (0.251 m)

force per unit length = 4.01×10⁻⁵ HA²/m²

Now let's just simplify the units to check that HA²/m² is equal to N/m. If you search for "Henry unit" you'll find that H=kgm²/s²A² so

HA²/m² = (kgm²/s²A²)(A²/m²) = kgm²A²/s²A²m² = kg/s²

and if you search for "Newton unit" you'll find that N=kgm/s², so N/m = (kgm/s²)/m = kg/s² which is the same as what we just showed HA²/m² is equal to

So the answer is force per unit length = 4.01×10⁻⁵ N/m

Simon

Answer 2

The force will be one of attraction.

If the wires are assumed to be 1 meter in length we get Force per Unit length from the formula

F=µI1I2/2πd = 4 x 10^-5 N