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 accelaration `3a_(0)` towards west. The electric and the maximum possible magnetic field in the room
(i) `(ma_(0))/(e)`, towards west
(ii) `(2 ma_(0))/(ev_(0))`, downward
(iii) `(ma_(0))/(e)`, towards east
(iv) `(2ma_(0))/(ev_(0))`, upward

Options

A. `(ma_(0))/(e )"west" (2ma_(0))/(ev_(0))` up
B. `(ma_(0))/(e ) "west" (2"ma"_(0))/(ev_(0))` down
C. `(ma_(0))/(e ) "East" (3"ma"_(0))/("ev"_(0))`up
D. `(ma_(0))/(e )"East" (3ma_(0))/("ev"_(0))` down

Answer

B

Explanation

Initial accelearation `a_(0)=(eE)/(m)`
`rarr E=(a_(0)m)/(e )`
`therefore (ev_(0)B_+eE)/(m)=3a_(0)`
or `ev_(0)B+eE=3a_(0)m`
`therefore ev_(0)B=3a_(0)-eE`
`rarr =3 ma_(0)-ma_(0)`
`rarr ev_(0)B=2 ma_(0)`
`therefore B=(2ma_(0))/(ev_(0))`