A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby losses one half of its kinetic energy. The radius of curvature of its path becomes :

Question

A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby losses one half of its kinetic energy. The radius of curvature of its path becomes : (A)  (1/√2)  (B)  √2  (C)  (3/√2)  (D)  (2/√3)

Tardigrade Admin · Accepted Answer

The radius of circular path is given by

r = \frac{μυ}{qB}

If

E_{K}

be kinetic energy of particle then its momentum

p = m υ = 2 m E_{K}

So, the radius is given by

r = \frac{2 m E _{K}}{qB}

So,

r \propto E_{K}

Therefore, when kinetic energy is halved the radius is reduced to

\frac{1}{2}

times its initial value.

Q. A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby losses one half of its kinetic energy. The radius of curvature of its path becomes :

$\frac{1}{2}$

$2$

$\frac{3}{2}$

$\frac{2}{3}$

Q. A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby losses one half of its kinetic energy. The radius of curvature of its path becomes :

2​1​

2​

2​3​

3​2​

The radius of circular path is given by r=qBμυ​ If EK​ be kinetic energy of particle then its momentum p=mυ=2mEK​​ So, the radius is given by r=qB2mEK​​​ So, r∝EK​​ Therefore, when kinetic energy is halved the radius is reduced to 2​1​ times its initial value.

$\frac{1}{2}$

$2$

$\frac{3}{2}$

$\frac{2}{3}$