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A proton moving withh a velocity 2.5xx10...

A proton moving withh a velocity `2.5xx10^(7)m//s` enters a magnetic field of intensity `2.5T` making an angle `30^(@)` with the magnetic field. The force on the proton is

A

`3 xx 10^(-12)` N

B

`5 xx 10^(-12)` N

C

`6 xx 10^(-12)` N

D

`9 xx 10^(-12) N`

Text Solution

Verified by Experts

The correct Answer is:
B
`F = qv B sin theta`
`= 1.6 xx 10^(-19) xx 2.5 xx 10^(7) xx 2.5 xx (1)/(2) = 5 xx 10^(-12) N`
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