When a proton is released from rest in a room, it starts with an initial acceleration `a_(0)` towards west. When it is projected towards north with a speed `v_(0)` it moves with an initial acceleration `3a_(0)` towards west. The electric and magnetic fields in the room are:

Options

A. `(ma_(0))/(e)" east, "(3ma_(0))/(ev_(9))" down"`
B. `(ma_(0))/(e)" west, "(2ma_(0))/(ev_(0))" up"`
C. `(ma_(0))/(e)" west, "(2ma_(0))/(ev_(0))" down"`
D. `(ma_(0))/(e)" east, "(3ma_(0))/(ev_(0))" up"`

Explanation

To solve the problem, we need to analyze the forces acting on the proton in two different scenarios: when it is released from rest and when it is projected towards the north with a speed \( v_0 \).

### Step 1: Analyze the first scenario (proton released from rest)
When the proton is released from rest, it experiences an initial acceleration \( a_0 \) towards the west. This indicates that there is an electric force acting on the proton in the westward direction.

- **Force due to Electric Field**:
\[ F_e = qE \]
where \( q \) is the charge of the proton and \( E \) is the electric field. Since the proton is positively charged, the direction of the electric force is the same as the direction of the electric field.

- **Acceleration**:
\[ F = ma \implies qE = ma_0 \implies E = \frac{ma_0}{q} \]

### Step 2: Analyze the second scenario (proton projected towards north)
In the second scenario, when the proton is projected towards the north with a speed \( v_0 \), it experiences an initial acceleration of \( 3a_0 \) towards the west.

- **Total Force**: The total force acting on the proton is the sum of the electric force and the magnetic force. The magnetic force is given by:
\[ F_m = q(v \times B) \]
Here, \( v \) is the velocity of the proton (which is towards the north), and \( B \) is the magnetic field.

- **Direction of Forces**:
Since the proton is moving north and the acceleration is towards the west, the magnetic force must also act towards the west to account for the increased acceleration.

### Step 3: Set up the equations
From the second scenario, we can express the net force acting on the proton:
\[ F_{net} = F_e + F_m \]
Given that \( F_e = qE \) and \( F_m = q(v_0 \times B) \), we can write:
\[ ma_{net} = qE + q(v_0 \times B) \]
Substituting the known values:
\[ m(3a_0) = qE + q(v_0 \times B) \]

### Step 4: Relate the two scenarios
From the first scenario, we know:
\[ E = \frac{ma_0}{q} \]
Substituting this into the second scenario:
\[ m(3a_0) = q\left(\frac{ma_0}{q}\right) + q(v_0 \times B) \]
This simplifies to:
\[ 3ma_0 = ma_0 + q(v_0 \times B) \]
Rearranging gives:
\[ 2ma_0 = q(v_0 \times B) \]

### Step 5: Solve for the magnetic field \( B \)
We can solve for \( B \):
\[ B = \frac{2ma_0}{qv_0} \]

### Conclusion
The electric field \( E \) is directed towards the west, and the magnetic field \( B \) is perpendicular to the velocity of the proton (which is north). Thus, the electric field is responsible for the initial acceleration \( a_0 \) towards the west, and the magnetic field contributes to the increased acceleration \( 3a_0 \) when the proton is moving north.