# Answer to Question #33279 in Electromagnetism for tarunika

Question #33279

a charged particle moving in a perpendicular uniform magnetic field penetrates a layer of head and there by looses half of its kinetic energy . how will the radius of its path change?

Expert's answer

If the kinetic energy was reduced 2 times, then velocitywas reduced sqrt(2) times, because

E=mv^2/2

the radius is related to velocity as

r = mv/qB

where m is mass, B is magnetic field and q is charge. If all these values stay the same, then change in radius is also in sqrt(2) times, as velocity is, because it is proportional to it.

Answer: radius will become sqrt(2) times smaller.

E=mv^2/2

the radius is related to velocity as

r = mv/qB

where m is mass, B is magnetic field and q is charge. If all these values stay the same, then change in radius is also in sqrt(2) times, as velocity is, because it is proportional to it.

Answer: radius will become sqrt(2) times smaller.

## Comments

## Leave a comment