In a region,10^{19} alpha-particles and 10^{1 | vedclass

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(C) The current $I$ is defined as the rate of flow of charge,$I = \frac{dq}{dt}$. By convention,the direction of current is the direction of flow of p

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$6.4 \, A$ towards left

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(C) The current $I$ is defined as the rate of flow of charge,$I = \frac{dq}{dt}$. By convention,the direction of current is the direction of flow of positive charge.$1$. For $\alpha$-particles: Each $\alpha$-particle has a charge of $+2e$. Since $10^{19}$ $\alpha$-particles move to the left per second,the current $I_{\alpha} = (10^{19} 	imes 2e) \, C/s$ towards the left.$2$. For protons: Each proton has a charge of $+e$. Since $10^{19}$ protons move to the left per second,the current $I_p = (10^{19} 	imes e) \, C/s$ towards the left.$3$. For electrons: Each electron has a charge of $-e$. Since $10^{19}$ electrons move to the right per second,the current $I_e$ is in the opposite direction of the electron flow,i.e.,towards the left. The magnitude is $I_e = (10^{19} 	imes e) \, C/s$ towards the left.Total current $I = I_{\alpha} + I_p + I_e = (2e + e + e) 	imes 10^{19} \, C/s = 4e 	imes 10^{19} \, C/s$.Substituting $e = 1.6 	imes 10^{-19} \, C$:$I = 4 	imes 1.6 	imes 10^{-19} 	imes 10^{19} \, A = 6.4 \, A$.Since all components contribute to current towards the left,the total current is $6.4 \, A$ towards the left.

In a region,$10^{19}$ $\alpha$-particles and $10^{19}$ protons per second move to the left,while $10^{19}$ electrons move to the right per second. The current is

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