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. The electric and magnetic fields in the room are:

• A. $\frac{m{a}_0}{e}east,\frac{3m{a}_0}{e{v}_0}down$
• B. $\frac{m{a}_0}{e}west,\frac{2m{a}_0}{e{v}_0}up$
• C. $\frac{m{a}_0}{e}west,\frac{2m{a}_0}{e{v}_0}down$
• D. $\frac{m{a}_0}{e}east,\frac{3m{a}_0}{e{v}_0}up$

Answer 1

Answer: (C)

Acceleration of charged particle... $\overline{a}=\frac{q}{m}(E+\overline{v}\times \overline{B})$
Released from rest $\Rightarrow \overline{a}=\frac{q}{m}\overline{E}=a_0(west)$ $\Rightarrow \overline{E}=\frac{m{a}_0}{e}(west)$
when it is projected towards north, acceleration due to magnetic force $2a_0$ Therefore magnetic field =$\frac{2m{a}_0}{e{v}_0}(down)$