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)` east, `(3ma_0)/(ev_0)` down
B. `(ma_0)/(e)` west, `(2ma_0)/(ev_0)` up
C. `(ma_0)/(e)` west, `(2ma_0)/(ev_0)` down
D. `(ma_0)/(e)` east, `(3ma_0)/(ev_0)` up

Answer

C

Explanation

In this question, the proton moves from rest towards west. It is due to a force on proton by virtue of electric field along west.
Acceleration of proton due to electric field
`=(eE)/(m)=a_0` or `E=(ma_0)/(e)` west
When proton is projected towards north with a speed `v_0`, it moves with an acceleration `3a_0` towards west, shows that the proton is experiencing forces due to electric field along west and magnetic field acting vertically downwards. Therefore, the acceleration of proton due to magnetic field `=3a_0-a_0=2a_0`.
Force on proton due to magnetic field
`m(2a_0)=ev_0` or `B=(2ma_0)/(ev_0)` downwards
Choice (c) is correct.