A proton released from rest in a room experiences an initial acceleration a₀ directed towards the west. If, instead, it is projected towards the north at a speed v₀ , its initial acceleration becomes 3 a₀ towards the west. Determine the electric field and the minimum possible magnetic field present in the room.

Step-by-step solution

Let the directions be represented by unit vectors: East as i , West as - i , North as j , South as - j , Upward as k , and Downward as - k . When the proton is released from rest, its velocity v = 0 . The only force acting on it is due to the electric field E . F _e = e E = m a e E = m(a₀(- i )) E = ma₀ e (- i ) Thus, the electric field is ma₀ e towards the west. When the proton is projected towards the north with speed v₀ , its velocity is v = v₀ j . The total force is the vector sum of the electric and magnetic f