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. The electric and magnetic fields in the room are:

Options

A. `(ma_(0))/(e)"east" , (3 ma_(0))/(ev_(0)) "down"`
B. `(ma_(0))/(e) "west" , (2ma_(0))/(ev_(0)) "up"`
C. `(ma_(0))/(e) "west" , (2ma_(0))/(ev_(0)) "down"`
D. `(ma_(0))/(e) "east" , (3ma_(0))/(ev_(0)) "up"`

Answer

C

Explanation

Initially `a_(0) = (eE)/(m)` (west) `implies E = (ma_(0))/(e)""` (west)
when projected towards north `(ev_(0) B + eE)/(m) = 3a_(0)`
`implies ev_(0) B + eE = 3 a_(0) m`
`ev_(0) = 3ma_(0) - eE = 2 ma_(0)`
`B = (2ma_(0))/(ev_(0))` (vertically downward direction)