Question

Two parallel very long straight wires carrying currents of 20 A and 30 A respectively are at a separation of 3 m between them. If the currents are in the same direction, find the attractive force between them per unit length.

Answer 1

Answer:

the attractive force between them is electromagnetic

Answer 2

Answer:

F = 40.127 × $10^{-6}$ N

Explanation:

We have been given ,

Two parallel current carrying current is I₁ = 20 A , I ₂ = 20 A

Separation of D = 3 m

attractive length L = 1 m

Now ,

We know that ,

Force between parallel wire is F =$\dfrac{\mu\ I1*I2*L}{2*\pi*D}$

     Therefore we know      μ =  1 . 26 ×$10^{-6}$ T m / A

F = $\dfrac{1.26 * 10^{-6}*20*30*1}{2*\pi* 3}$

F = 40.127 × $10^{-6}$ N