A proton is projected with a speed of `3X10^6 ms ^(-1)` horizontally from east to west. A uniform magnetic field `vec B` of strength `2.0X10^(-3) T` exists in the vertically upward direction(a) find the force on the proton just after it is projected. (b) what is the acceleration produced?

Explanation

(a) The situation is shown in figure . The force is
perpendicular to `vecB` hence it is in the horizontal plane
through the proton. In theis place, it is perpendicular to
the velocity `vec v`. Thus, it is along the north-south line. The
rule for vector product shows that `vec v xx vec B` is towards north.
As the charge on the proton is positive, the force
`vec F = vec qu X vec B` is also towardas north. the magnitude of the
force is
`F= qv B sintheta`
`= ((1.6 xx 10^(-19)C) (3.0 xx10^6 ms^(-1) (2.0 xx 10^(-3)T)`
`= 9.6 xx10^(-16)N`
. (b) The acceleration of the proton is
`a = (F/m)= (9.6xx10^(-16) N)/1.67 xx 10^(-27)` kg
` =5.8 xx 10^11 ms^(-2)`.