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 accelaration `3a_(0)` towards west. The electric and the maximum possible magnetic field in the room
(i) `(ma_(0))/(e)`, towards west
(ii) `(2 ma_(0))/(ev_(0))`, downward
(iii) `(ma_(0))/(e)`, towards east
(iv) `(2ma_(0))/(ev_(0))`, upward

Options

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

Answer

To solve the problem, we will analyze the two scenarios provided in the question step by step.

### Step 1: Analyze the first scenario
When the proton is released from rest, it experiences an initial acceleration \( a_0 \) towards the west. Since the proton is positively charged, it will accelerate in the direction of the electric field.

1. **Force on the proton**: The force acting on the proton can be expressed as:
\[ F = ma_0 \]
where \( m \) is the mass of the proton.

2. **Electric field**: The force due to the electric field can be expressed as:
\[ F = qE \]
where \( q \) is the charge of the proton and \( E \) is the electric field.

3. **Setting the forces equal**: Since both forces are acting on the proton, we can set them equal:
\[ ma_0 = qE \]

4. **Solving for the electric field**: Rearranging gives us:
\[ E = \frac{ma_0}{q} \]
Since the charge of the proton \( q \) is denoted as \( e \), we can write:
\[ E = \frac{ma_0}{e} \]
The direction of the electric field is towards the west.

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

1. **Net force on the proton**: The net force acting on the proton can be expressed as:
\[ F_{\text{net}} = m(3a_0) \]

2. **Forces acting on the proton**: In this case, the forces acting on the proton are due to both the electric field and the magnetic field. The magnetic force can be expressed as:
\[ F_B = q(v_0 \times B) \]
Since the angle between the velocity \( v_0 \) (north) and the magnetic field \( B \) is 90 degrees, we can simplify this to:
\[ F_B = qv_0B \]

3. **Setting up the equation for forces**: The total force acting on the proton can be expressed as:
\[ F_{\text{net}} = F_E + F_B \]
Substituting the expressions for the forces gives:
\[ m(3a_0) = ma_0 + qv_0B \]

4. **Rearranging the equation**: We can rearrange this to find the magnetic field \( B \):
\[ m(3a_0) - ma_0 = qv_0B \]
\[ 2ma_0 = qv_0B \]

5. **Solving for the magnetic field**: Rearranging gives:
\[ B = \frac{2ma_0}{qv_0} \]
Substituting \( q \) with \( e \) (the charge of the proton):
\[ B = \frac{2ma_0}{ev_0} \]

### Step 3: Determine the direction of the magnetic field
Using the right-hand rule, we can determine the direction of the magnetic field. The velocity is towards the north, and the net force (due to acceleration) is towards the west. Therefore, the magnetic field must be directed downward.

### Conclusion
From the analysis, we find:
1. The electric field \( E \) is given by:
\[ E = \frac{ma_0}{e} \quad \text{(towards west)} \]
2. The magnetic field \( B \) is given by:
\[ B = \frac{2ma_0}{ev_0} \quad \text{(downward)} \]

### Final Answers:
- The electric field is \( \frac{ma_0}{e} \) towards west (Option i).
- The magnetic field is \( \frac{2ma_0}{ev_0} \) downward (Option ii).