When a proton is released from rest in a room, it starts with an initial acceleration a₀ towards west. When it is projected towards north with a speed v₀, it moves with an initial acceleration 3a₀ towards west. The electric field and the maximum possible magnetic field in the room
(i) $\frac{m a_{0}}{e}$ , towards west
(ii) $\frac{2 m a_{0}}{e v_{0}}$ , downward
(iii) $\frac{m a_{0}}{e}$ , towards east
(iv) $\frac{2 m a_{0}}{e v_{0}}$ , upward

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(i), (iv)

(ii), (iv)

(ii), (iii)

(i), (ii)

answer is A.

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Detailed Solution

Initially

$F_{1} = {ma}_{0}$

${ma}_{0} = qE$

$E = {ma}_{0} / q Along west$

In case of projection with velocity $v_{0}$ , magnetic force also acts on the particle perpendicular to the direction of projection

$\begin{array}{l} F_{1} + F_{2} = 3 {ma}_{0} \\ {ma}_{0} + F_{2} = 3 {ma}_{0} \end{array}$

Maximum magnetic force is

$F_{2} = {qV}_{0} B$

$\begin{array}{l} So, {qV}_{0} B = 2 {ma}_{0} \\ B = 2 {ma}_{0} / {qV}_{0} (down). \end{array}$

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