A rod 0.45 m long slides at 2.4 m/s along two rails in a field of 0.30 T, at right angles to the…

Step 1. The emf. $e = Blv = 0.30\ \text{T} \times 0.45\ \text{m} \times 2.4\ \text{m/s} = 0.324\ \text{V}$.

Step 2. The current. $I = \dfrac{e}{R} = \dfrac{0.324\ \text{V}}{0.54\ \Omega} = 0.6\ \text{A}$.

Step 3. The force. $F = BIl = 0.30\ \text{T} \times 0.6\ \text{A} \times 0.45\ \text{m} = 0.081\ \text{N}$. It acts against the motion, so the pull must be $0.081$ N along the motion.

Step 4. The power. $P = Fv = 0.081\ \text{N} \times 2.4\ \text{m/s} = 0.194\ \text{W}$.