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 $3 a_{0}$ toward west. The electric and magnetic fields in the room are

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$\frac{m a_{0}}{e} e a s t, \frac{3 m a_{0}}{e v_{0}} d o w n$

$\frac{m a_{0}}{e} w e s t, \frac{2 m a_{0}}{e v_{0}} d o w n$

$\frac{m a_{0}}{e} e a s t, \frac{3 m a_{0}}{e v_{0}} u p$

$\frac{m a_{0}}{e} w e s t, \frac{2 m a_{0}}{e v_{0}} u p$

answer is D.

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Detailed Solution

A stationary proton experiences a force along the direction of electric fi eld. Given, acceleration of the proton is a0 along west, so ${M a}_{0} = q E$
$E = \frac{M a_{0}}{q} along west.$
When it is projected towards north, it experiences an accleration 3a₀ along west. Acceleration produced by magnetic fi eld is 2a₀, if north is taken as $\hat{j}$ , west is along $- \hat{i}$
$\begin{array}{l} 2 m a_{0} (- i) = q (v_{0} \hat{j} \times \vec{B}) \\ | \vec{B} | = \frac{2 m a_{0}}{e v_{0}} \end{array}$
direction must be $(- \hat{k})$ , i.e., downward.
So, $E = \frac{{ma}_{0}}{e} west$
$B = \frac{2 {m a}_{0}}{{e v}_{0}} downward$