Initial acceleration =a0
when a proton is released from rest in a room it starts with an initial acceleration a0 towards west.
F1= electric force acting on the proton in west direction
q= charge on proton
E= electric field in the region in west direction
m= mass of proton
electric force on the proton can be given on F1=qE
Using Newton's second law
F1=ma0
hence,
ma0=qE
E=ma0/q Along west
Case -2
V0= initial velocity of the particle
F2= magnetic force acting on the proton in west direction electric force on proton =3ma0
F1+F2=3ma0
ma0+F2=3ma0
maximum magnetic force is given by
F2=qV0B
So, qV0B=2ma0
B=2ma0/qV0 (down).
Option C