Q122 / 180 · NEET 2013 (All India) · Physics

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

  • A
    $\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ up
  • Correct answer
    B
    $\frac{m a_0}{e}$ west, $\frac{2 m a_0}{e v_0}$ down
  • C
    $\frac{m a_0}{e}$ east, $\frac{3 m a_0}{e v_0}$ up
  • D
    $\frac{m a_0}{e}$ east $\frac{3 m a_0}{e v_0}$ down

Solution

Initial acceleration, $a_0=\frac{\partial E}{m}$ $\begin{aligned} & \Rightarrow \quad E=\frac{a_0 m}{e} \therefore \frac{e v_0 B+e E}{m}=3 a_0 \\ & \text { or } \quad e v_0 B+e E=3 a_0 m \\ & \therefore \quad e v_0 B=3 m a_0-e E \\ & \Rightarrow \quad=3 m a_0-m a_0 \quad[\text { fromeq. (i)] } \\ & \Rightarrow \quad e v_0 B=2 m a_0 \\ & \therefore \quad B=\frac{2 m a_0}{e v_0} \\ & \end{aligned}$

Hence, the correct answer is option: 'B'.

Relevant Questions