Question

Two long, straight wires are parallel and 5 cm apart. One carries a current of 2.0 A, the other a current of 5.0 A. If the two currents flow in the same directions, what is the magnitude and direction of the force per unit length of one wire on the other?

Answer 1

F/L = 4×10-⁵N/m (force per unit length)

Step-by-step explanation:

Given I1 = 2.0A, I2 = 5.0A, r = 5.0cm = 5×10-²m

To find: F/L

F/L = μo×I1×I2/2πr

μo =4π×10-⁷T.m/A

r = distance between both wires.

F/L = 4π×10-⁷×2.0×5.0/(2π×5×10-²)

F/L = 4×10-⁵N/m.

This force is attractive as the wires attract each other. Two wires(conductor ) carrying current in the same direction attract each other those that carry current in opposite directions repel each other.

Answer 2

The force unit length of one wire on the other is 4.0 x 10⁻⁵ N/m

Step-by-step explanation:

Given;

distance between the two wires, r = 5 cm = 0.05 m

current on the first wire, I₁ = 2A

current on the second wire, I₂ = 5A

Force per unit length on two parallel current carrying conductors is given as:

`(F)/(L) = (\mu_oI_1I_2)/(2\pi r)`

substitute in the values given into the equation;

`(F)/(L) = (4\pi *10^(-7) *2*5)/(2\pi *0.05) = 4.0 *10^(-5) \ N/m`

Therefore, the force unit length of one wire on the other is 4.0 x 10⁻⁵ N/m