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))`, upwardA. `(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" `

When a proton is released from rest in a room, it starts with an initi

1.	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))`, upwardA. `(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» Correct Answer - 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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