State the microscopic current magnitude relation involving \(n, |q|, A,\) and \(v_d\).

In a metal, is electron drift in the same direction as conventional current or the opposite direction?

If electron drift speed increases while \(n\) and \(A\) stay fixed, what happens to current magnitude?

A wire has \(n=8.0\times10^{28}\,\mathrm{m^{-3}}\), \(A=1.0\times10^{-6}\,\mathrm{m^2}\), \(v_d=2.0\times10^{-4}\,\mathrm{m\,s^{-1}}\), and electron carriers. Find the current magnitude.

A \(2.0\,\mathrm{A}\) current flows in a wire with \(n=5.0\times10^{28}\,\mathrm{m^{-3}}\) and \(A=2.5\times10^{-6}\,\mathrm{m^2}\). Estimate electron drift speed.

For positive charge carriers with \(n=1.0\times10^{24}\,\mathrm{m^{-3}}\), \(q=+2e\), and \(\vec v_d=3.0\times10^{-5}\hat\imath\,\mathrm{m\,s^{-1}}\), find \(\vec J\).

A metal wire narrows to half its original area while carrying the same steady current. Assuming carrier density is unchanged, by what factor does drift speed change in the narrow section?

Using \(\vec J=nq\vec v_d\), explain why negative carriers drifting left can produce conventional current to the right.

A conductor has two types of mobile carriers: electrons with density \(n_e\) and drift velocity \(\vec v_e\), and positive ions with density \(n_i\), charge \(+e\), and drift velocity \(\vec v_i\). Derive the total current density.

A wire carries steady current \(I\), but its area varies slowly as \(A(x)\). Assuming one carrier type with constant \(n\), derive \(v_d(x)\) and explain why a smaller area does not imply charge is accumulating there in steady state.