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: eE = ma0 towards west
⇒ E = ma0e, directed west (the proton is positive)
When moving north with v0, the total force is 3ma0 towards west
⇒ magnetic force FB = 3ma0 - ma0 = 2ma0 towards west
ev0B = 2ma0 ⇒ B = 2ma0ev0
Direction: take east = i, north = j, up = k. FB = e v B must be along - i with v along j.
j (- k) = - i ⇒ B is along - k, i.e. vertically down
∴ E = ma0e west, B = 2ma0ev0 down
Answer: (4)