Two horizontal rails are 60 cm apart, and a rod slides along them at 2.5 m/s…

Step 1. The emf. The vertical part of the field is $B\sin 30^\circ = 0.80\ \text{T} \times 0.50 = 0.40$ T. So $\varepsilon = (B\sin 30^\circ)\,lv = 0.40\ \text{T} \times 0.60\ \text{m} \times 2.5\ \text{m/s} = 0.600\ \text{V}$.

Step 2. The current. $I = \dfrac{\varepsilon}{R} = \dfrac{0.600\ \text{V}}{0.50\ \Omega} = 1.20\ \text{A}$.

Step 3. The force along the rails. The magnetic force on the rod has the size $BIl = 0.80\ \text{T} \times 1.20\ \text{A} \times 0.60\ \text{m} = 0.576\ \text{N}$. It is at right angles to the field. Its part along the rails is $BIl\sin 30^\circ = 0.288$ N, and this part opposes the motion. You must pull with $0.288$ N.

Step 4. The power. $P = Fv = 0.288\ \text{N} \times 2.5\ \text{m/s} = 0.720\ \text{W}$. The electrical power is $\varepsilon I = 0.600\ \text{V} \times 1.20\ \text{A} = 0.720$ W. They agree.