In a neon discharge tube,2.9 times 10^{18} Ne | vedclass

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(B) The current $i$ is defined as the rate of flow of charge. When both positive and negative charges move in opposite directions,their currents add u

Vedclass · Accepted Answer

$0.66 \, A$ towards right

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(B) The current $i$ is defined as the rate of flow of charge. When both positive and negative charges move in opposite directions,their currents add up in the direction of the positive charge flow.Let $n_+$ be the number of positive ions moving to the right per second and $n_-$ be the number of electrons moving to the left per second.The current due to positive ions is $i_+ = \frac{n_+ e}{t} = 2.9 	imes 10^{18} 	imes 1.6 	imes 10^{-19} \, A = 0.464 \, A$ (towards the right).The current due to electrons is $i_- = \frac{n_- e}{t} = 1.2 	imes 10^{18} 	imes 1.6 	imes 10^{-19} \, A = 0.192 \, A$ (towards the right,because electrons moving to the left constitute a current to the right).The total current $i = i_+ + i_- = 0.464 \, A + 0.192 \, A = 0.656 \, A \approx 0.66 \, A$ towards the right.

In a neon discharge tube,$2.9 \times 10^{18}$ $Ne^+$ ions move to the right each second,while $1.2 \times 10^{18}$ electrons move to the left per second. The electron charge is $1.6 \times 10^{-19} \, C$. The current in the discharge tube is:

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