The intensity of the magnetic force F experienced by a charge q moving with speed v in a magnetic field of intensity B is equal to
[tex]F=qvB \sin \theta[/tex]
where [tex]\theta[/tex] is the angle between the directions of v and B.
1) Re-arranging the previous formula, we can calculate the value of the magnetic field intensity. The charge is [tex]q=2.7 \mu C=2.7 \cdot 10^{-6}C[/tex]. In this case, v and B are perpendicular, so [tex]\theta=90^{\circ}[/tex], therefore we have:
[tex]B= \frac{F}{qv \sin \theta} = \frac{5.75 \cdot 10^{-3}N}{(2.7 \cdot 10^{-6}C)(773m/s)\sin 90^{\circ}}=2.8 T [/tex]
2) In this second case, the angle between v and B is [tex]\theta=55^{\circ}[/tex]. The charge is now [tex]q=42.0 \mu C=42.0 \cdot 10^{-6}C[/tex], and the magnetic field is the one we found in the previous part, B=2.8 T, so we can find the intensity of the force experienced by this second charge:
[tex]F=qvB \sin \theta=(42\cdot 10^{-6}C)(1.21 \cdot 10^3 m/s)(2.8 T)(\sin 55^{\circ})=0.12 N[/tex]
