In a discharge tube, `6 xx 10^18` electrons move to the left and `4 xx 10^18` protons move to the right past a cross-section in every second. Find the magnitude and the direction of current.

Explanation

Electrons :
`i_e = (q)/(t) = (n_e e)/(t)`
`i_p = (q)/(t) = (n_p e)/(t)`
Total current `i = i_e + i_p = ((n_e + n_p)e)/(t)`
=`(10 xx 10^18 xx 1.6 xx 10^-19)/(1)`
=`1.6 A`, towards right.
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