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. (i), (ii)
B. (ii), (iv)
C. (ii), (iii)
D. (i), (iv)

Answer

A

Explanation

`F = ma_(0)`
`eE = m a_(0) rArr E = (m a_(0))/(e)`, along west
Total acceleration is `3 a_(0), 2 a_(0)` is due to magnetic force

`F_(m) = m . 2 a_(0)`
`B e v_(0) = 2 m a_(0)`
`B = (2 ma_(0))/(ev_(0))`
`-2ma_(0)hat(i) = e[(v_(0) hat(j)) xx (- B hat(k))]`
`B` is in downward direction.