when a proton is released from rest in a room, it starts with an initial acceleration `a_0` towards west . When it is projected towards north with a speed `v_0,` it moves with an initial acceleration `3a_0` towards west. Find the elecrtic field and the minimum possible magnetic field in the room.

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)