An electron is moving at a speed of 3.2×10⁷ m s⁻¹ in a magnetic field of 12×10⁻⁴ T perpendicular to the direction of motion of the electron. The radius of the path of the electron is cm. (e=1.6×10⁻¹⁹ C and mₑ=9×10⁻³¹ kg)

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(3) 15. Given: v=3.2×10⁷ m s⁻¹, B=12×10⁻⁴ T perpendicular to the velocity, e=1.6×10⁻¹⁹ C, mₑ=9×10⁻³¹ kg. Idea: the magnetic force supplies the centripetal force, evB=(mₑv²)/r, so r=(mₑv)/(eB). Top: 9×10⁻³¹×3.2×10⁷=2.88×10⁻²³. Bottom: 1.6×10⁻¹⁹×12×10⁻⁴=1.92×10⁻²². r=(2.88×10⁻²³)/(1.92×10⁻²²)=0.15 m=15 cm.

An electron is moving at a speed of 3.2×10⁷ m s⁻¹ in a magnetic field of 12×10⁻⁴ T perpendicular to the direction of motion of the electron. The radius of the path of the electron is ______ cm. (e=1.6×10⁻¹⁹ C and mₑ=9×10⁻³¹ kg)

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