A square frame of side 13 cm and resistance 0.058 Ω is pushed at a steady 1.7 m/s through a region…

Step 1. The current while entering. $e = Bav = 0.21\ \text{T} \times 0.13\ \text{m} \times 1.7\ \text{m/s} = 0.0464\ \text{V}$. So $I = \dfrac{0.0464\ \text{V}}{0.058\ \Omega} = 0.800\ \text{A}$.

Step 2. The force. $F = BIa = 0.21\ \text{T} \times 0.800\ \text{A} \times 0.13\ \text{m} = 0.0218\ \text{N}$, against the motion.

Step 3. The heat in one stage. The stage lasts $\dfrac{a}{v} = \dfrac{0.13\ \text{m}}{1.7\ \text{m/s}} = 0.0765\ \text{s}$. The heat is $I^2Rt = (0.800\ \text{A})^2 \times 0.058\ \Omega \times 0.0765\ \text{s} = 2.84\ \text{mJ}$.

Step 4. The whole crossing. There are two stages, entering and leaving. $Q = 2 \times 2.84\ \text{mJ} = 5.68\ \text{mJ}$.