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A charged particle is moving on circular...

A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

A

8 times

B

4 times

C

2 times

D

16 times

Text Solution

Verified by Experts

The correct Answer is:
B
As `Bqv=(mv^(2))/(r) or r=(mv)/(Bq)`
According to the question, `v'=2v and B'(B)/(2)`
`therefore" "r'=(mv')/(B'q)=(m(2v))/((B//2)q)=(4mv)/(Bq)=4r`
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