Question

An electron and a proton, moving side by side at the same speed, enter a 0.020-T magnetic field. The electron moves in a circular path of radius9.0 mm. What is the radius of the proton?

Answer 1

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,

$F_e = q_e vB$.........(1)

$F_p = q_p vB$

using newton second law

$F = m a = m\dfrac{v^2}{r}$

equating Force due to electron and proton

$F_e = F_p$

$\dfrac{m_ev^2}{r_e}=\dfrac{m_pv^2}{r_p}$

$\dfrac{m_e}{r_e} = \dfrac{m_p}{r_p}$

m_ e = 9.1 x 10⁻³¹ Kg    and m_p = 1.67 x 10⁻²⁷ Kg

$r_p = \dfrac{m_p}{m_e}\times r_e$

$r_p = \dfrac{1.67\times 10^{-27}}{9.1 \times 10^{-31}}\times 9 \times 10^{-3}$

$r_p = 16.5 m$

Hence, the radius of proton is equal to 16.5 m.