Answer:
[tex]8.4\cdot 10^{-13} N[/tex]
Explanation:
The magnitude of the magnetic force on the proton is given by:
[tex]F=qvB sin \theta[/tex]
where:
[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge
[tex]v=4.2\cdot 10^6 m/s[/tex] is the proton velocity
[tex]B=2.5 T[/tex] is the magnetic field
[tex]\theta=30^{\circ}[/tex] is the angle between the direction of v and B
Substituting into the formula, we find
[tex]F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N[/tex]
