Question

Two long, straight, parallel wires of length 3.7 m carry parallel currents of 3.6 A and 1.6 A. (a) If the wires are separated by a distance of 3.1 cm, what is the magnitude of the force between the two wires

Answer 1

`|F|=1.37* 10^(-4)\ N`

Step-by-step explanation:

Given:

Length of the parallel wires (L) = 3.7 m

Current in the first wire (I₁) = 3.6 A

Current in the second wire (I₂) = 1.6 A

Separation between the parallel wires (d) = 3.1 cm = 0.031 m [1 cm = 0.01 m]

Both the wires will attract each with forces equal in magnitude and opposite in direction.

Now, the magnitude of the force acting between the two wires is given as:

`|F|=(\mu_0I_1I_2L)/(2\pi d)`

Where,
`\mu_0\to permeability\ constant=4\pi * 10^(-7)\ N/A^2`

Plug in the given values and solve for |F|. This gives,

`|F|=((4\pi* 10^(-7)\ N/A^2)(3.6\ A)(1.6\ A)(3.7\ m))/(2\pi* 0.031\ m)\\\\|F|=1.37* 10^(-4)\ N`

Therefore, the magnitude of the force between the two wires is
`|F|=1.37* 10^(-4)\ N`