Through a give cross section `n_(1)` electrons per second are passing from left to right and `n_(2)` protons per second are passing from right to left simultaneously. The electric current throught that cross section is ( `theta` = electronic charge).

Options

A. `(n_(1) + n_(2))`e towards left
B. `(n_(2) - n_(1))` e toward right
C. `(n_(1)^(2) + n_(2)^(2))` e toward left
D. `(n_(2)^(2) - n_(1)^(2))` e toward right

Answer

To solve the problem, we need to calculate the total electric current through a given cross-section when `n1` electrons are moving from left to right and `n2` protons are moving from right to left.

### Step-by-step Solution:

1. **Understanding the Charge Carriers**:
- Electrons are negatively charged particles and are moving from left to right.
- Protons are positively charged particles and are moving from right to left.

2. **Direction of Current**:
- By convention, the direction of electric current is taken as the direction of positive charge movement. Therefore:
- The current due to electrons (moving left to right) will be considered as flowing in the opposite direction (right to left).
- The current due to protons (moving right to left) will be considered as flowing in the same direction (right to left).

3. **Calculating Current from Electrons**:
- The charge carried by one electron is denoted as `e`.
- If `n1` electrons are passing per second, the total charge carried by the electrons per second (which is the current due to electrons) is:
\[ I_e = n1 \cdot e \]
- Since the current due to electrons is in the direction from right to left, we will consider this as a positive contribution to the total current.

4. **Calculating Current from Protons**:
- Similarly, if `n2` protons are passing per second, the total charge carried by the protons per second (which is the current due to protons) is:
\[ I_p = n2 \cdot e \]
- The current due to protons is also in the direction from right to left, so this will also be a positive contribution to the total current.

5. **Total Current Calculation**:
- The total current `I` through the cross-section is the sum of the currents due to electrons and protons:
\[ I = I_e + I_p = n1 \cdot e + n2 \cdot e \]
- Simplifying this, we get:
\[ I = (n1 + n2) \cdot e \]

6. **Direction of Total Current**:
- Since both contributions are in the same direction (right to left), the total current is also directed from right to left.

### Final Answer:
The electric current through that cross-section is:
\[ I = (n1 + n2) \cdot e \quad \text{(towards left)} \]