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

Answer

B

Explanation

`vecF_(L)=vecF_(e)+vecF_(m)=qvecE+q(vecvxxvecB)`
`ma_(0)=qE+0 " " ` ...(1)
`E=(ma_(0))/(q) " so "vecE=(ma_(0))/(e)` in west
`vecF_(L)=vecF_(e)+vecF_(m)`
`3ma_(0)=qE+q(vecvxxvecB)rArr q(vecvxxvecB)=2ma_(0)`(west)
`vecF_(m)=qvecvxxvecB`
`-hati=(+hatj)xx.......`
`B=(2ma_(0))/(qv)` in vertically downward