The magnetic field produced by a current carrying wire is:

`B=(mu_0 I)/(2pi r)=(mu_0 I)/(2pi d)`

Then use the lorentz force law per unit length is

`F/L=q/L(v xx B)=lambda (v xx B)=(lambda*v xx B)=I xx B=I*(mu_0 I)/(2pi d) `

`F/L=(mu_0 I^2)/(2pi d)`

Now lets find the direction of the force. I...

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The magnetic field produced by a current carrying wire is:

`B=(mu_0 I)/(2pi r)=(mu_0 I)/(2pi d)`

Then use the lorentz force law per unit length is

`F/L=q/L(v xx B)=lambda (v xx B)=(lambda*v xx B)=I xx B=I*(mu_0 I)/(2pi d) `

`F/L=(mu_0 I^2)/(2pi d)`

Now lets find the direction of the force. I will use cylindrical coordinates `(r,phi,z)` . Let the current go in the `z ` direction. Then the magnetic field will wrap around the wire in the `phi` direction by the right hand rule. Now lets look at the cross product.

`F=I xx B=z xx phi=-r`

Therefore, the magnetic force on the other wire is directed radially inward or toward the wire. You would find the same answer for the other wire. Hence the magnetic force is **attractive** for wires with currents in the same direction.

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