Answer:
16.5 m
Explanation:
Given,
Magnetic field = 0.02 T
radius of electron = 9 mm
speed of electron = speed of proton
radius of proton = ?
We know,
[tex]F_e = q_e vB[/tex].........(1)
[tex]F_p = q_p vB[/tex]
using newton second law
[tex]F = m a = m\dfrac{v^2}{r}[/tex]
equating Force due to electron and proton
[tex]F_e = F_p[/tex]
[tex]\dfrac{m_ev^2}{r_e}=\dfrac{m_pv^2}{r_p}[/tex]
[tex]\dfrac{m_e}{r_e} = \dfrac{m_p}{r_p}[/tex]
m_ e = 9.1 x 10⁻³¹ Kg and m_p = 1.67 x 10⁻²⁷ Kg
[tex]r_p = \dfrac{m_p}{m_e}\times r_e[/tex]
[tex]r_p = \dfrac{1.67\times 10^{-27}}{9.1 \times 10^{-31}}\times 9 \times 10^{-3}[/tex]
[tex]r_p = 16.5 m[/tex]
Hence, the radius of proton is equal to 16.5 m.