When a proton is released from rest in a room, it starts with an initial acceleration $a_0$ towards west.…
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 $3 a_0$ towards west. The electric and magnetic fields in the room are
$\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ up
$\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ down
$\frac{m a_0}{e}$ east, $\frac{3 m a_0}{e v_0}$ up
$\frac{m a_0}{e}$ east $\frac{3 m a_0}{e v_0}$ down
Solution
Initial acceleration, $a_0=\frac{\partial E}{m}$
$\begin{aligned} & \Rightarrow \quad E=\frac{a_0 m}{e} \therefore \frac{e v_0 B+e E}{m}=3 a_0 \\ & \text { or } \quad e v_0 B+e E=3 a_0 m \\ & \therefore \quad e v_0 B=3 m a_0-e E \\ & \Rightarrow \quad=3 m a_0-m a_0 \\ & \Rightarrow \quad e v_0 B=2 m a_0 \\ & \therefore \quad B=\frac{2 m a_0}{e v_0} \\ & \end{aligned}$