Question

A charged particle moving in a uniform magnetic field penetrates a layer of lead and loses one-half of its kinetic energy. The radius of curvature changes to:

Answer 1

radius of curvature is given as, $r=\frac{mv}{qB}$

• here m is mass of particle
• v is velocity .
• q is charge.
• B is magnetic field.

we know, Kinetic energy, K= 1/2 mv²

or, 2K = mv²

or, 2Km = (mv)²

or, √{2Km} = mv

so, radius of curvature will be , r = √{2Km}/qB,

hence, it is clear that, radius of curvature is directly proportional to square root of kinetic energy.

a/c to question,

final kinetic energy , K' = 1/2 K

so, radius of curvature will be ;

$\frac{r}{r_f}=\sqrt{\frac{K}{\frac{1}{2}K}}$

or, $r_f=\frac{r}{\sqrt{2}}$