This field is uniform along wire 2 and perpendicular to it, and so the force F_{2} it exerts on wire 2 is given by F = IlB sin\theta with sin \theta = 1: \[F_{2} = I_{2}lB_{1}.\label{22.11...This field is uniform along wire 2 and perpendicular to it, and so the force F_{2} it exerts on wire 2 is given by F = IlB sin\theta with sin \theta = 1: F_{2} = I_{2}lB_{1}.\label{22.11.2} By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F_{2}. (Note that F_{1} = -F_{2}.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length.