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A charged particle moving in a uniform magnetic field penetrates a layer of lead and there by loses one-half of its kinetic energy. How does the radius of curvature of its path change?

Text Solution

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Kinetic energy, `K=1/2mv^2` or `v=sqrt((2K)/(m))`.
Radius of curvature,
`r=(mv)/(qB)=(m)/(qB)sqrt((2K)/(m))=(sqrt(2mK))/(qB)`
Therefore, `rpropsqrtK`,
So `r^'/r=sqrt((K//2)/(K))=1/sqrt2` or `r^'=r/sqrt2`
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