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

Question

A charged particle moving in a uniform magnetic field penetrates the layer of lead and thereby losses one-half of its kinetic energy. The radius of curvature of its path (A) The radius reduces to r√2 (B) The radius reduces to (r/√2)  (C) The radius remains the same (D) The radius becomes r/2

Tardigrade Admin · Accepted Answer

r= (√2 m E/q B) and r 1=√(2 textitm (E â¡ / 2)/q B) So, r 1 = (r â¡/√2)

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

The radius reduces to $r 2$

The radius reduces to $\frac{r}{2}$

The radius remains the same

The radius becomes $r /2$

$r = \frac{2 m E}{qB}$ and $r_{1} = \frac{2 m ( E /2 )}{qB}$
So, $r_{1} = \frac{r}{2}$

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

The radius reduces to r2​

The radius reduces to 2​r​

The radius remains the same

The radius becomes r/2

r=qB2mE​​ and r1​=qB2m(E⁡/2)​​ So, r1​=2​r⁡​

The radius reduces to $r 2$

The radius reduces to $\frac{r}{2}$

The radius becomes $r /2$

$r = \frac{2 m E}{qB}$ and $r_{1} = \frac{2 m ( E /2 )}{qB}$
So, $r_{1} = \frac{r}{2}$