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A proton enters a magnetic fiedl of flux...

A proton enters a magnetic fiedl of flux density `1.5"weber"//m^(2)` with a velocity of `2xx10^(-7)m//sec` at an angle of `30^(@)` with the field. The force on the proton will be

A

`2.4 xx 10^(-12) N`

B

`0.249 xx 10^(-12) N`

C

`24 xx 10^(-12) N`

D

`0.0024 xx 10^(-2)N `

Text Solution

Verified by Experts

The correct Answer is:
A
As, `F = q vB sin theta`
`rArr R = (1.6 xx 10^(-19)) xx 2 xx 10^(7) xx 1.5 xx sin 30^(@)`
` rArr = F = 1.6^(-19) xx 2 xx 10^(7) xx 1.5 xx (1)/(2)`
`rArr = 2.4 xx 10^(-12) N`
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