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)` toward west. The electric and magnetic fields in the room are

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 step by step, we will analyze the forces acting on the proton in both scenarios described in the question.

### 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.

- Since the proton is at rest initially, the only force acting on it must be due to an electric field, as a magnetic field does not exert a force on a stationary charge.
- The force acting on the proton can be expressed as:
\[ F = m a_0 \]
where \( m \) is the mass of the proton.

### Step 2: Determine the electric field
The force due to the electric field \( E \) acting on the proton can be given by:
\[ F = qE \]
where \( q \) is the charge of the proton (which is equal to the elementary charge \( e \)).

Setting the two expressions for force equal gives:
\[ ma_0 = qE \]
Thus, the electric field \( E \) can be expressed as:
\[ E = \frac{ma_0}{q} \]
Since the proton is positively charged, the direction of the electric field is also towards the west.

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

- In this case, both the electric field and the magnetic field will exert forces on the proton.
- The total force acting on the proton can be expressed as:
\[ F_{\text{total}} = F_{\text{electric}} + F_{\text{magnetic}} = 3ma_0 \]

### Step 4: Express the forces
The electric force remains the same as before:
\[ F_{\text{electric}} = qE = ma_0 \]
The magnetic force can be expressed as:
\[ F_{\text{magnetic}} = qvB \]
where \( B \) is the magnetic field strength.

### Step 5: Set up the equation for forces
Substituting the expressions for the forces into the total force equation gives:
\[ ma_0 + qvB = 3ma_0 \]
Rearranging this, we find:
\[ qvB = 3ma_0 - ma_0 = 2ma_0 \]

### Step 6: Solve for the magnetic field \( B \)
Now, substituting \( q = e \) (the charge of the proton) into the equation:
\[ e v_0 B = 2ma_0 \]
Solving for \( B \) gives:
\[ B = \frac{2ma_0}{ev_0} \]

### Step 7: Determine the direction of the magnetic field
To find the direction of the magnetic field, we can use the right-hand rule:
- The proton is moving north (upward in our coordinate system).
- The magnetic force is directed towards the west.

Using the right-hand rule, if the thumb points in the direction of the velocity (north), and the fingers point in the direction of the magnetic force (west), then the palm will face downwards, indicating that the magnetic field \( B \) is directed downwards.

### Summary of Results
- The electric field \( E \) is directed towards the west with a magnitude of \( \frac{ma_0}{e} \).
- The magnetic field \( B \) is directed downwards with a magnitude of \( \frac{2ma_0}{ev_0} \).