1.

A proton entersa magnetic field of 1.5 T with a velocity of 2 xx 10^(7) ms^(-1) at an angle of 30^@ with the field. The force on the proton will be

Answer»

`2.4 xx 10^(-12) N`
`0.24 xx 10^(-12) N`
`24 xx 10^(-12)N`
`0.024 xx 10^(-12) N`

Solution :Here `q = +E = 1.6 xx 10^(-19) C, B = 1.5 T , v = 2 xx 10^(7) m s^(-1) and theta = 30^@`
`:.` Force on PROTON `F = q v B sin theta = 1.6 xx 10^(-19) xx 2 xx 10^(7) xx 1.5 xx sin 30^@ = 2.4 xx 10^(-12) N`.


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