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In the previous problem, the maximum ran...

In the previous problem, the maximum range of movement of the centre of the part of the circle from line AD in which charged particle of charge Q moves with a velocity v when `(theta)` is positive to when `(theta)` is negative is given by

A

`+-(mv)/(2QB)`

B

`+-(mv)/(QB)`

C

`+-(2mv)/(QB)`

D

`+-(2mv)/(3QB)`

Text Solution

Verified by Experts

The correct Answer is:
B
An observation will show that when `theta` is positive and actually `90^(@)` angle formed will be `2 pi` at the center and centre of the circle formed by the charged particle will be at a distance of radius of the circle from AD.
It will be also true for `theta` to be negative. (`theta` measured anticlockwise form normal AC).
Hence range is ` +- r = +-(mv)/(QB)`.
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