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A charge having q/m equal to 10^(8) c/kg...

A charge having q/m equal to `10^(8)` c/kg and with velocity `3 xx 10^(5)` m/s enters into a uniform magnetic field B = 0.3 tesla at an angle `30^(@)` with direction of field. Then radius of curvature will be:

A

0.01cm

B

0.5cm

C

1cm

D

2cm

Text Solution

Verified by Experts

The correct Answer is:
D
Particle moving in a circular path the required centripetal force is provided by the magnetic force. Therefore, `evB sin theta= (mv^(2))/(R )`
`rArr R= (mv)/(eB sin theta)= (3 xx 10^(5))/(10^(8) xx 0.3 xx 1//2) =2cm`
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