First slide

Force on a charged particle moving in a magnetic field

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

Moderate

A

twice the original radius
B

$\sqrt{2}$ times the original radius
C

half the original radius
D

(1 / $\sqrt{2}$ ) times the original radius

Solution

The radius of curvature of a charged particle moving in uniform magnetic field is given by
$\begin{array}{l} r = \frac{mv}{qB} = \frac{\sqrt{2 mK}}{qB} \\ r^{'} = \frac{\sqrt{2 {mK}^{'}}}{qB} \\ ∴ \frac{r^{'}}{r} = \sqrt{(\frac{K^{'}}{K})} = \sqrt{\frac{K / 2}{K}} = \frac{1}{\sqrt{2}} \\ r^{'} = \frac{1}{\sqrt{2}} \cdot r \end{array}$

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