MCQ

**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)`

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#### 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)`.

Concept: The Earth’s Magnetism

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