A single charged ion has a mass of 1.13× 10^ -23 . It is accelerated through a potential difference of 500 V . It enters a magnetic field of 0.4 T , moving perpendicular to the field . The radius of its path in the field is ?
Answers
Answer:
20
Explanation:
I hope you understand
Given: mass of charged ion= 1.13× 10^ -23, Potential difference= 500V, Magnetic field= 0.4T
To find: Radius of the path of charged ion in the field.
Solution: With the help of radial acceleration and Newton's second law of motion, the magnetic field can be find through the given formula:
B= mv/qR
where, m is mass of the particle, v is speed of the particle, q is charge on the particle, R is radius of the path.
We know that, the charged particle is accelerating through a potential difference.
Therefore work done can be written as
eV= 1/2mv^2
v=√(2eV/m)
putting the values of e, V, m in the above equation we will get v that is speed of the ion
v= √( 2× (1.6×10^-19C)×(500V)/(1.13×10^-23)
v= 3.75× 10^3 m/s
we now have speed of the particle, putting in the formula B= mv/qR, we will get R
0.4= 1.13×10^-23×3.75×10^3/(1.6×10^-19C)× R
R= 0.662m
Therefore, the radius of the path of charged particle in the field will be R= 0.662m.