A wire 0.25 m long carries 4 A in a field of 0.35 T…

Step 1. The largest force. $BIL = 0.35\ \text{T} \times 4\ \text{A} \times 0.25\ \text{m} = 0.35\ \text{N}$.

Step 2. (a) Along the field, $\sin 0^\circ = 0$: $F = 0\ \text{N}$.

Step 3. (b) $F = 0.35\ \text{N} \times 0.5 = 0.175\ \text{N}$.

Step 4. (c) $F = 0.35\ \text{N} \times 1 = 0.35\ \text{N}$.