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 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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