Question

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. Find the electric field and the minimum possible magnetic field in the room.

Answer 1

When proton released from rest if gets ao acceleration towards west,

i.e., qE = $m{a}_0$

$\Rightarrow E=\frac{m{a}_0}{q}$ to wards west.

When it projected towards north with velocity $v_0$, then force on proton

$\underset{F}{\to }=q\underset{v_0}{\to }+\underset{B_0}{\to }\left|\underset{F}{\to }\right|=q{v}_{0}B\Rightarrow B=\frac{\left|\underset{F}{\to }\right|}{q{v}_0}$

Acceleration towards west due to force

$=3a_0-a_0=2a_0$

Hence required magnetic field

$B=\frac{2m{a}_0}{q{v}_0}$