Question

A charged particle moving in a uniform magnetic field penetrates a layer of lead and their by loses one half of its kinetic energy how does the radius of curvature of its path change

Answer 1

Answer:

1√2 rg.

Explanation:

Since, we know that the the gyroradius or the radius of curvature of a given charged particle (also known as radius of gyration, Larmor radius or cyclotron radius) is given by the formulae of :  rg = m*v(perpendicular)/qB.

From the above equation we get that the m is the mass of the given particle, v is the component of the perpendicular velocity along the direction of the uniform magnetic field, and the q is the electric charge of the particle, while B is the magnetic field strength of the uniform magnetic field.

Since, the the particle looses half of the kinetic energy but still it somehow did not scatter in the different trajectory, hence-:

v^2perpendicular ,after=1/2*v^2⊥.

Hence, v perpendicular,after=1√2v perpendicular.

And so, radius of curvature will be =1√2 rg.