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When a proton is released from rest in a...

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

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" `

Text Solution

Verified by Experts

The correct Answer is:
B
`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
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