Answer:
[tex] \underline{ \boxed{F_m = 3.35 { \times 10}^{ - 2} N}}[/tex]
Explanation:
[tex]since \: in \: this \: case \to \: \underline{ v }\: is \perp \: to \:\underline{ q} \\ then \: the \: magnetic \: force \: (F_m) \: is \: max = qvB \sin(90) \\ F_m = qvB \times (1) = qvB \\ hence \to \\ F_m = qvB \\ where \\ \: q = 0.75 C : v = 235m {s}^{ - 1} : B = 1.9 { \times 10}^{ - 4} T \\ therefore \to \\ F_m = 0.75 \times 235 \times 1.9 { \times 10}^{ - 4} \\ F_m = 176.25 \times 1.9 { \times 10}^{ - 4} \\ F_m = 0.0334875 = 3.35 { \times 10}^{ - 2} N \\ \boxed{F_m = 3.35 { \times 10}^{ - 2} N}[/tex]
