Question

A charged particle, moving at right angle to a uniform magnetic field, starts moving along a circular arc of radius of curvature $r$. In the field it now penetrates a layer of lead and loses ${\frac{3}{4}}^{th}$ of it's initial kinetic energy. The radius of curvature of it's path now will be :

• A. $4r$
• B. $2r$
• C. $\frac{r}{4}$
• D. $\frac{r}{2}$

Answer 1

The correct option is D $\frac{r}{2}$

Initially, $r=\frac{mv}{qB}$
$r=\frac{\sqrt{2K_{e}m}}{qB}$
Now ${K_e}^{\prime }=\frac{1}{4}{K}_e$
$r^{\prime }=\frac{\sqrt{2\times (\frac{1}{4})K_{e}m}}{qB}$
=$\frac{r}{2}$