A charged particle moving at right angles to a uniform magnetic field and starts moving along acircular arc of radius of curvature ‘r’. In the field it now penetrates a layer of lead and loses 3/4th of its initial kinetic energy. The radius of curvature of its path now will be

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r/4

r/2

answer is D.

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Detailed Solution

$K E_{2} = K E_{1} - \frac{3}{4} K E_{1} = \frac{K E_{1}}{4}; K E \propto r^{2} \Rightarrow \frac{K E_{2}}{K E_{1}} {(\frac{r_{2}}{r_{1}})}^{2} \Rightarrow r_{2} = \frac{r_{1}}{2}$

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