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A charged particle of charge q and mass m, gets deflected through an angle `theta` upon passing through a square region of side 'a' which contains a uniform magnetic field B normal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle ?

A

`(qB)/(m) "" a cot theta`

B

`(qB)/(m) "" a tan theta`

C

`(qB)/(m) "" cot^(2) theta`

D

`(qB)/(m) ""a tan^(2) theta`

Text Solution

Verified by Experts

The correct Answer is:
A
Now sin `theta = (a)/( R) , R = (mv)/(qB)`
`therefore v = (qBa cot theta)/(m)`
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