A metal rod 30 cm long slides at 3.0 m/s along two rails, in a field of 0.45 T at right angles to…

Step 1. The emf. $\varepsilon = Blv = 0.45\ \text{T} \times 0.30\ \text{m} \times 3.0\ \text{m/s} = 0.405\ \text{V}$.

Step 2. The current. $I = \dfrac{\varepsilon}{R} = \dfrac{0.405\ \text{V}}{0.90\ \Omega} = 0.450\ \text{A}$.

Step 3. The force. $F = BIl = 0.45\ \text{T} \times 0.450\ \text{A} \times 0.30\ \text{m} = 0.0608\ \text{N}$, that is $60.8$ mN.

Step 4. The power. $P = Fv = 0.0608\ \text{N} \times 3.0\ \text{m/s} = 0.182\ \text{W}$, that is $182$ mW.