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Moving Charge in a Magnetic Field

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

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Answer: (d)

At rest only the electric force acts: $eE = ma_0$ towards west, so $\vec{E} = \dfrac{ma_0}{e}$ west.
Moving north, the total is $3ma_0$ west, so the magnetic force is $2ma_0$ west: $ev_0B = 2ma_0 \Rightarrow B = \dfrac{2ma_0}{ev_0}$ .
With east $= \hat{i}$ , north $= \hat{j}$ , up $= \hat{k}$ : $\hat{j} \times (-\hat{k}) = -\hat{i}$ (west), so $\vec{B}$ points down.
So the answer is (d) $\frac{ma_0}{e}$ west, $\frac{2ma_0}{ev_0}$ down.

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