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When a proton is released from rest in a room, it starts with an initial acceleration a0towards west. When it is projected towards north with a speed υ0

it moves with an initial acceleration 3a0 toward west. The electric and magnetic fields in the room are

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a
ma0eeast,3ma0eυ0up
b
ma0eeast,3ma0eυ0down
c
ma0ewest,2ma0eυ0up
d
ma0ewest,2ma0eυ0down

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detailed solution

Correct option is D

when proton is stationary force is only due to electric field  i.e, E⇀ =ma0e along westforce due to magnetic field must be   2ma0along west  its velocity is along north (j) force is along  west (-i)F→=q(V ⇀ xB⇀)    (-i=j x-k)  hence magnetic field is  vertically downwards direction and having a magnitude 2ma0e

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