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Question
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 2` r
4r
`r/sqrt 2`
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Solution
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 `bbunderline(sqrt 2 r)`.
Explanation:
When a charged particle is accelerated through a potential difference V,
`1/2 mv^2` = qv
⇒ v = `sqrt((2 q V)/m)`
Radius in magnetic field:
r = `(m v)/(q B)`
Since v ∝ `sqrt V`,
∴ r ∝ `sqrt V`
If V is doubled,
r' = `sqrt 2 r`
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