 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ib/paper-3/17d/b/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

The Lorentz force on the current-carrying bar is

F = I L e_{y} \times B_{0} e_{z} = \frac{B _{0} L}{R} (E_{0} - B_{0} L \overset{s}{˙}) e_{x} .

There is no friction force along the rails. The battery part accelerates the bar rightward for positive

E_{0} B_{0}

, whereas the induced-current part is the opposing force

- B_{0}^{2} L^{2} \overset{s}{˙} / R

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