Answer
0.358 T
Work Step by Step
Use equation 20–2. The length of the wire in the field is the diameter of the pole face (i.e., the approximate length where the field envelops the wire).
$$F_{max}=I\mathcal{l}B $$
$$B=\frac{F_{max}}{I\mathcal{l} }=\frac{1.28N}{(0.555m)(6.45A)}=0.358T$$
