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{ma_0}{e} east, \frac{3ma_0}{ev_0} down$
• B. $ \frac{ma_0}{e} west, \frac{2ma_0}{ev_0} up$
• C. $ \frac{ma_0}{e} west, \frac{2ma_0}{ev_0} down$
• D. $ \frac{ma_0}{e} east, \frac{3ma_0}{ev_0} up$

Answer 1

Answer: (C)

Acceleration of charged particle... $\overline a = \frac {q}{m} \, (E + \overline v\times \overline B)$
Released from rest $\Rightarrow \overline a = \frac{q}{m}\overline E = a_0 (west)$ $\Rightarrow \overline E = \frac{ma_0}{e} (west)$
when it is projected towards north, acceleration due to magnetic force $2a_0$ Therefore magnetic field =$\frac{2ma_0}{ev_0}(down)$