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Through a give cross section n(1) elec...

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).

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

Text Solution

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To solve the problem, we need to determine the total electric current through a given cross-section where `n1` electrons are moving from left to right and `n2` protons are moving from right to left. ### Step-by-Step Solution: 1. **Understanding the Direction of Current:** - Electrons have a negative charge and are moving from left to right. However, the conventional current direction is opposite to the flow of electrons. Therefore, the current due to electrons will flow from right to left. - Protons have a positive charge and are moving from right to left. The current due to protons will also flow from right to left. 2. **Calculating Current Due to Electrons:** - The current due to electrons can be calculated using the formula: \[ I_e = n1 \cdot e \] where \( n1 \) is the number of electrons passing per second and \( e \) is the charge of an electron. 3. **Calculating Current Due to Protons:** - The current due to protons can be calculated similarly: \[ I_p = n2 \cdot e \] where \( n2 \) is the number of protons passing per second. 4. **Total Current Calculation:** - Since both currents (due to electrons and protons) are flowing in the same direction (from right to left), we can add them together: \[ I_{total} = I_e + I_p = n1 \cdot e + n2 \cdot e \] - This simplifies to: \[ I_{total} = (n1 + n2) \cdot e \] 5. **Final Result:** - The total current through the cross-section is: \[ I = (n1 + n2) \cdot e \] - The direction of the current is from right to left.
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