Chapter 22 - Magnetism - Problems and Conceptual Exercises - Page 792: 12

$2.2\times 10^{-5}N$

We know that $\frac{F_{New}}{F_{old}}=\frac{qv_{new}Bsin\theta_{new}}{qv_{old}Bsin\theta_{old}}=\frac{v_{new}sin\theta_{new}}{v_{old}sin\theta_{old}}$ We plug in the known values to obtain: $\frac{F_{New}}{F_{old}}=\frac{(6.3)sin25^{\circ}}{(27)sin90^{\circ}}=0.099$ $F_{new}=0.099F_{old}=0.099(2.2\times 10^{-4})=2.2\times 10^{-5}N$