ma₀ west, 2ma₀v₀ down
ma₀ west, 2ma₀v₀ up
ma₀ east, 3ma₀v₀ down
ma₀ west, 3ma₀v₀ up
The force on a proton due to an electric field E is given by F = qE, where q is the charge of the proton. The acceleration a is given by Newton's second law, F = ma, where m is the mass of the proton. Therefore, the electric field is related to the acceleration by:
a = qE/m
Given that the initial acceleration is a₀ towards north, the electric field in the north direction is:
E_north = ma₀/q
When the proton moves with speed v₀, an additional acceleration of 3a₀ towards west is observed. This acceleration is due to the magnetic force, which is given by F = qvB, where B is the magnetic field. The magnetic force is perpendicular to both the velocity and the magnetic field.
Therefore, we have:
3a₀ = qv₀B/m
Solving for B:
B = 3ma₀/(qv₀)
The direction of the magnetic field can be determined using the right-hand rule. Since the force is towards west and the velocity is towards north, the magnetic field must be directed downwards.
Therefore, the electric field is ma₀/q towards north and the magnetic field is 3ma₀/(qv₀) downwards. However, the options provided only give the fields in the east-west and up-down directions. Let's re-examine the initial acceleration a₀ to the north. This is caused by the electric field alone. The x-component of the acceleration a₀ = 0. The y-component is a₀. The electric field in the y-direction is E_y = ma₀/q.
The additional acceleration of 3a₀ is only in the x-direction due to the magnetic force. This is given by F = qv x B, where v = v₀j and F = -3ma₀i. Thus, -3ma₀i = q(v₀j) x B. By cross product, B must point into the page (in the -z direction) to produce a force in the -x direction. However, the given options only offer E in east-west and B in up-down. The solution doesn't directly match any option. There might be a discrepancy in the question or options provided.