when a proton is released from rest in a room, it starts with an initital 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. Find the elecrtic field and the minimum possible magnetic field in the room.

Options

A. `(ma_(0))/(e)` west, `(2ma)/(e v_(0))` up
B. `(ma_(0))/(e)` west, `(2ma)/(e v_(0))` down
C. `(ma_(0))/(e)` east, `(3ma_(0))/(e v_(0))` up
D. `(ma_(0))/(e)` east, `(3ma_(0))/(e v_(0))` down

Answer

B

Explanation

When the proton is relasted from rest, it will experience force only due to electric field. Initial accelearation of proton `a_(0) = (eE)/(m) implies (ma_(0))/(e) 0`
Now the proton si projected towards north with a speed `v_(0)`
Initial acceleration of proton `3a_(0) = (ev_(0) B + eE)/(m)`
`implies ev_(0) + eE = 3 ma_(0)`
`implies ev_(0)B = 3ma_(0) - eE = 3 ma_(0) - ma_(0) = 2ma_(0)`
`implies B = (2ma_(0))/(ev_(0))`