The figure below shows two parallel cylindrical copper wires that carry the
same current of I in the same direction.
Refer to the Image
The two wires are a distance D = 7 cm apart and the radius of each wire is
R = 0.1 mm. Determine I if the total force on the bottom wire is zero and
that weight acts downward. Assume that the top wire is held in place and
that the density of copper is δ = 8920 kg/m3 and μ0 = 4π × 10−7 T·m/A.
The force per unit length between the two wires both carrying current I is determined by the formula
`f = F/l = (mu_0*I^2)/(2pid)` ,
where d is the distance between the wires, d = 7 cm = 0.07 m.
The total force on the bottom wire will thus depend on the wire's length l, which is not given:
`F = (mu_0*I^2)/(2pid) l`
The two wires carrying current in the same direction attract each other, so this force is directed upward. The gravitational force on the wire is directed downward. Since the total force on the bottom wire is zero, the magnitudes of the magnetic and the gravitational force must be equal.
The gravitational force equals mg, where m is the mass of the wire. The mass can be found using the density of the copper and the radius of the wire,
R = 0.1 mm = 0.0001 m
The volume of the cylindrical wire is
`V = piR^2*l`
and the mass is
`m = sigma*V = sigma*pi*R^2*l =`
= `8920*pi*(0.0001)^2*l = 2.8*10^(-4)*l`
Setting magnetic force equal to gravitational, we get
`(mu_0*I^2)/(2pid) l = 2.8*10^(-4) *l * g`
The length of the wire, l, cancels. Now, after plugging the values of the constants and d = 0.07 m, the current I can be found:
`(4pi*10^(-7)*I^2)/(2pi*0.07) = 2.8*10^(-4)*9.8`
`I^2 = 960.4 A^2`
`I = 30.99 A` , or approximately 31 Amperes.
The current in the wires is 31 Amperes.
According to Ampere's law, the magnetic forces between the two conductors are equal and opposite and have the following expression:
F = μ0 I1I2 l/2Π R
Where R is the distance between the wires, I represents the current and l is the length. Whenever currents moving in the same direction, the forces between the wires are of attraction
The bottom wire will be at rest, ie the total force on it is zero, if the magnetic force acting on it is equal to its weight; then we have:
μ0 I˄2 l/2Π R = mg (1)
In this equation, we have written I˄2, because both currents are equal
To assess the length l in this equation, we will use the density and note that the wires are cylindrical. The density is the mass per unit volume:
ρ = m/V
The volume of a wire is:
V = Π r˄2 l
Where r is the radius of the wire cross section, and l is the length. Then substituting in the equation for density, and solving the length l we have:
l = m/Π r˄2 ρ
Evaluating the length l in the equation (1) and solving for I˄2:
I˄2 = 2Π˄2 R r˄2 ρ g/ μ0
Taking the square root of both sides:
I = 0.83 A
The actual value of the current is:
I = 30.99 A