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Question

When a proton is released from rest in 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 and magnetic fields in the room are :
  1. ma0eeast,3ma0ev0down
  2. ma0ewest,3ma0ev0up
  3. ma0eeast,3ma0ev0up
  4. ma0ewest,2ma0ev0down

A
ma0eeast,3ma0ev0down
B
ma0eeast,3ma0ev0up
C
ma0ewest,3ma0ev0up
D
ma0ewest,2ma0ev0down
Solution
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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

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