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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
Detailed Solution
$\vec a = \frac{q}{m}(\vec E + \vec v\times\vec B)$
Released from rest: $\vec a = \frac{q}{m}\vec E = a_0$ (west) $\Rightarrow E = \frac{ma_0}{e}$ (west)
Projected north, the acceleration due to the magnetic force $= 3a_0 - a_0 = 2a_0$ (west)
$ev_0B = 2ma_0 \Rightarrow B = \frac{2ma_0}{ev_0}$ (down)
Released from rest: $\vec a = \frac{q}{m}\vec E = a_0$ (west) $\Rightarrow E = \frac{ma_0}{e}$ (west)
Projected north, the acceleration due to the magnetic force $= 3a_0 - a_0 = 2a_0$ (west)
$ev_0B = 2ma_0 \Rightarrow B = \frac{2ma_0}{ev_0}$ (down)