A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loses half…

A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loses half of its kinetic energy, then the radius of curvature of its path is
  1. No change
  2. Reduced by $\frac{1}{2}$ times of its initial values
  3. Reduced to $\frac{1}{\sqrt{2}}$ times of its initial values
  4. Reduce to $\frac{1}{4}$ times of its initial values

Solution

We know that $r=\frac{m V}{q B}=\frac{\sqrt{2 m(K . E)}}{q B}$ $\begin{aligned} & \Rightarrow \mathrm{r} \propto \sqrt{\mathrm{K} . \mathrm{E}} \\ & \Rightarrow \frac{\mathrm{r}_2}{\mathrm{r}_1}=\sqrt{\frac{\mathrm{K} \cdot \mathrm{E}_2}{\mathrm{~K} \cdot \mathrm{E}_1}}=\sqrt{\frac{1}{2}}=\frac{1}{\sqrt{2}} \Rightarrow \mathrm{r}_2=\frac{\mathrm{r}_1}{\sqrt{2}}\end{aligned}$

Asked in: AP EAMCET 2022 (05 Jul Shift 2)

Practice more Magnetic Effects of Current questions on Aicharya