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)` toward west. The electric and magnetic fields in the room are

Options

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

Answer

a

Explanation

`ma_(0) = e E rArr E = (ma_(0))/e ` (towards west ) and `(ev_(0)B + eE)/m = 3 a_(0)`
`rArr ev_(0) B = 3ma_(0) - eE = 3 ma_(0) = ma_(0) = 2 ma_(0)`
` rArr B = (2 ma_(0))/(ev_(0)) ` (downwards)