Question

A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field directed perpendicular to the path of the proton. What is the radius of the proton’s resulting orbit?

Answer 1

Answer:

0.114m

Explanation:

From the general expression for the radius of the proton's resulting orbit, we have

$r=\frac{mv}{qB}$

where q is is the charge of the proton $1.6*10^{-19}C$

m is the mass of the proton  $1.67*10^{-27}kg$

B is the magnetic field  $0.040T$

and v i the speed.

to determine the speed, we use the expression

Kinetic Energy=$qV$

$1/2mv^{2}=qV$

where V  is the voltage value i.e 1.0kv

and v is the speed

Hence, from simple rearrangement we have the speed v to be

$v=\sqrt{\frac{2Vq}{m}}$

if we substitute value, we have

$v=\sqrt{\frac{2*1000*1.6*10^{-19} }{1.67*10^{-27}}}$

carrying out careful arithmetic we arrive at

$v=4.38*10^{5} m/s$.

using the value for the speed in the expression for the radius of the orbit as stated earlier, we have

$r=\frac{1.67*10^{-27}*4.38*10^{5}}{1.6*10^{-19}*0.04}$

$r=0.114m$