Respuesta :
Answer:
a) [tex]v=6.898\times 10^{6}\ m.s^{-1}[/tex] is the speed of each proton
b) [tex]F_c=3.97\times 10^{-13}\ N[/tex]
Explanation:
Given:
radius of path of motion, [tex]r=0.2\ m[/tex]
we know charge on protons, [tex]q=1.6\times 10^{-19}\ C[/tex]
magnetic field strength, [tex]B=0.36\ T[/tex]
we've mass of proton, [tex]m=1.67\times 10^{-27}\ kg[/tex]
a)
From the equivalence of magnetic force and the centripetal force on the proton:
[tex]F_B=F_C[/tex]
[tex]q.v.B=\frac{m.v^2}{r}[/tex]
[tex]q.B=\frac{m.v}{r}[/tex]
where:
v = speed of the proton
[tex](1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}[/tex]
[tex]v=6.898\times 10^{6}\ m.s^{-1}[/tex] is the speed of each proton
b)
Now the centripetal force on each proton:
[tex]F_c=m.\frac{v^2}{r}[/tex]
[tex]F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}[/tex]
[tex]F_c=3.97\times 10^{-13}\ N[/tex]
