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Question
Choose the correct answer from given options
A charge particle after being accelerated through a potential difference 'V' enters in a uniform magnetic field and moves in a circle of radius r. If V is doubled, the radius of the circle will become
Options
2r
`sqrt(2r)`
4r
`r/sqrt(2)`
Solution
The relation between the accelerating potential and the accelerating voltage is given as: `r = sqrt(2mqV)/(qB)`
As the potential is doubled the radius of curvature becomes `sqrt(2)` times.
Hence, the correct answer is option `sqrt(2r)`.
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