Question

An unknown particle moves in a straight line through crossed electric and magnetic fields with e = 1.5 kv/m and b = 0.034 t. if the electric field is turned off, the particle moves in a circular path of radius r = 2.7 cm. what might the particle be

Answer 1

Answer:

Alpha particle

Explanation:

Initially, the particle moves in a straight line. This means that the electric force and the magnetic force are equal:

$qE=qvB$

where q is the charge, E is the electric field, v is the speed and B is the magnetic field.

In this problem,

E = 1.5 kV/m = 1500 V/m

B = 0.034 T

Solving the equation for v we find the speed of the particle

$v=\frac{E}{B}=\frac{1500 V/m}{0.034 T}=4.41\cdot 10^4 m/s$

Now the electric field is turned off, so the particle starts moving in a circular motion, where the magnetic force acts as centripetal force:

$qvB=m\frac{v^2}{r}$

where

r = 2.7 cm = 0.027 m is the radius of the circular path

Solving the problem for q/m, we find charge-to-mass ratio of the particle

$\frac{q}{m}=\frac{v}{Br}=\frac{4.41\cdot 10^4 m/s}{(0.034 T)(0.027 m)}=4.8\cdot 10^7 C/kg$

And this corresponds to the q/m ratio of an alpha particle, which has:

$q=2e=3.2\cdot 10^{-19}C\\m=4a.m.u.=6.64\cdot 10^{-27} kg$