What is the Current Generated by a Single Electron in Uniform Circular Motion?

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A single electron in uniform circular motion generates an average current, which can be calculated using the formula I = eω/(2π). Given an angular speed of ω = 2 x 10^16 rad/s and the charge of the electron e = 1.602 x 10^-19 C, the resulting current is approximately 5.1 x 10^-4 A. The radius of the circle does not affect the current calculation in this context. The current direction is counterclockwise (CCW). The discussion highlights the importance of including direction in current calculations.
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A single electron in uniform circular motion constitutes an average current around the circle. If the electron moves clockwise with angular speed in a circle of radius r, find the current. Evaluate I if \omega=2X10^{16} rad/s and r = 5 nm.

I never used radius r in my answer. This book usually doesn't thrown in irrelavant information in the questions. Did I do this right?

\begin{array}{l}<br /> I = \left| {\frac{Q}{t}} \right| \\ <br /> \\ <br /> t = \frac{d}{v} = \frac{{2\pi }}{\omega } \\ <br /> \\ <br /> I = \left| {\frac{Q}{{\left( {\frac{{2\pi }}{\omega }} \right)}}} \right| = \left| {\frac{{Q\omega }}{{2\pi }}} \right| = \left| {\frac{{ - e\omega }}{{2\pi }}} \right| = \frac{{e\omega }}{{2\pi }} \\ <br /> \\ <br /> \frac{{1.602 \times 10^{ - 19} {\rm{C}} \times 2 \times 10^{16} {\rm{rad/s}}}}{{2\pi }} = 5.1 \times 10^{ - 4} {\rm{C/s = }}5.1 \times 10^{ - 4} {\rm{A}} \\ <br /> \end{array}

 
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Right. The current would be in the CCW direction. (I haven't checked the arithmetical calculation.)
 
Thanks for verifying that, and reminding me to include direction.
 
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