Final answer:
A proton released from rest in a room accelerates due to an electric field E = ma0/e westwards. When projected north, its acceleration due to a magnetic field is B = 2ma0/ev0 upwards.
Step-by-step explanation:
When a proton is released from rest and experiences an initial acceleration a0 towards the west, it indicates the presence of an electric field in that direction. The force on the proton is due to this electric field and can be calculated using F = ma, where m is the proton's mass, a is its acceleration, and F is the force it experiences. Given that the charge of a proton is e, the electric field, E, can be derived using F = qE, leading to E = ma/e. Thus, when the proton is released from rest, the electric field is E = ma0/e towards the west.
When the proton is projected towards the north with speed v0 and experiences an initial acceleration of 3a0 towards the west, a magnetic field is also present. The additional acceleration is due to the magnetic Lorentz force, which is perpendicular to both the velocity of the proton and the magnetic field. Applying the right-hand rule, we can infer that the magnetic field is directed upwards. The magnetic force on a moving charge is F = qvB, yielding B = F/qv = 3ma0/ev0. Hence, the magnetic field is B = 2ma0/ev0 up.