Question

A) A proton is moving in a circular orbit of radius 13.5 cm in a uniform 0.325 T magnetic field directed perpendicular to the velocity of the proton. Find the translational speed of the proton. B) If an electron moves perpendicular to the same magnetic field with this same translational speed, what is the radius of its circular orbit?

Answer 1

Answer:

$r=8.97*10^{-5}m$

Explanation:

We need first make a couple of consideration.

Mass of proton ($m_p$)is $1.67*10^{-27}$

Mass of electron is ($m_e$) $9.11*10^{-31}$

charge of proton or electron(q)= $1.6*10^{-19}c$

So for

$r=13.5cm$

$B=0.325T$

Given the relation

$r=\frac{mv}{qB}$ (1)

So clearing v,

$v=\frac{qBr}{m} = \frac{(1.6*10^-{19})(0.325)(13.5*10^{-2})}{1.67*19^{-27}}$

$v= 5.1240*10^6m/s$

So with the equation (1)

$r=\frac{mv}{qB} = \frac{(9.11*10^{-31})(5.1240*10^6)}{(1.6*10^{-19})(0.325)}$

$r=8.97*10^{-5}m$