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A positively charged particle of charge ...

A positively charged particle of charge 'q' enters in a uniform magnetic field B directed inward and is deflected a distance x after travelling a distance Y as shown in the figure. The magnitude of linear momentum of the particle is:

A

`(qBY)/2`

B

`(qB)/2(x^2/Y+x)`

C

`(qB)/(2x)(x^2+Y^2)`

D

`(qB)/(2Y)(x^2+Y^2)`

Text Solution

Verified by Experts

The correct Answer is:
C
`R sin theta =Y`
`R(1-cos theta )=x`
`rArr (1-x/R)^2 + (Y/R)^2 =1 rArr R=1/(2x)(x^2+Y^2)`
`rArr p=mv=(qB)/(2x)(x^2+Y^2)`
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