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.
