When a proton is released from rest in a gravity free room, it starts with an initial acceleration `a_(0)` towards east.When it is projected towards south with a speed `v_(0)`, it moves with an initial acceleration `3a_(0)` towards east.The minimum possible magnetic field in the room is `(Nma_(0))/(ev_(0))`.Find the value of `N`.

Answer

B

Explanation

`vecF_(E)=ma_(0)` towards east
`+eE=ma_(0)`
`E=(ma_(0))/e` towards east.
Net acceleration `=3a_(0)`
acc due to electric force `=a_(0)` (along east)
acc due to megnetic force `=2a_(0)` (along east)
`vecF_(B)=2ma_(0)hati`
`qV_(0)B sin theta=2ma_(0)`
`B=(2ma_(0))/(qV_(0)sin theta)`
`B_(min)=(2ma_(0))/(qV_(0))`
For magnetic force along west with `theta=90^(@)`
Direction of magnetic field must be along `+z`.(upward)