An electron having 18.2 keV kinetic energy is moving on a circular path in a magnetic field of 10⁻⁴ T. The speed of the electron is . (mass of electron m=9.1×10⁻³¹ kg)

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(1) 8×10⁷ m s⁻¹. Given: K=18.2 keV=1.82×10⁴×1.6×10⁻¹⁹=2.912×10⁻¹⁵ J and m=9.1×10⁻³¹ kg. Idea: the magnetic force is always perpendicular to the velocity, so it does no work and the speed follows from the kinetic energy alone — the 10⁻⁴ T is not needed here. K=1/2mv²⇒ v=√(2K)/m. (2K)/m=(5.824×10⁻¹⁵)/(9.1×10⁻³¹)=6.4×10¹⁵ m² s⁻². v=√6.4×10¹⁵=8×10⁷ m s⁻¹.

An electron having 18.2 keV kinetic energy is moving on a circular path in a magnetic field of 10⁻⁴ T. The speed of the electron is ______. (mass of electron m=9.1×10⁻³¹ kg)

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