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 and magnetic fields in the room are
- ma0e east, 3ma0ev0 up
- ma0e east, 3ma0ev0 down
- ma0e west, 2ma0ev0 up
- ma0e west, 2ma0ev0 down
Answer
(4) ma0e west, 2ma0ev0 down
Solution
At rest only the electric force acts, so eE = ma0 towards west, which gives E = ma0e, directed west (the proton is positive).
When it moves north with v0, the total force is 3ma0 towards west, so the magnetic force is FB = 3ma0 - ma0 = 2ma0 towards west and
ev0B = 2ma0 ⇒ B = 2ma0ev0
For the direction, take east = i, north = j, up = k. FB = e v B must be along - i with v along j.
Since j (- k) = - i, B is along - k, i.e. vertically down.
∴ E = ma0e west, B = 2ma0ev0 down
Answer: (4)