An electron with kinetic energy 1.50 keV circles in a place perpendicular to a uniform magnetic field. The orbit radius is 30.0 cm. Find the speed of…

Question

An electron with kinetic energy 1.50 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 30.0 cm. Find the speed of the electron and the magnitude of the magnetic field B. (m_e = 9.11 × 10⁻³¹ kg, q_e = 1.6 × 10⁻¹⁹ C, 1eV = 1.6 × 10⁻¹⁹ J)

Answer 1

An electron with kinetic energy 1.50 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 30.0 cm. Find the speed of the electron and the magnitude of the magnetic field B. (m_e = 9.11 × 10⁻³¹ kg, q_e = 1.6 × 10⁻¹⁹ C, 1eV = 1.6 × 10⁻¹⁹ J)

3.2 × 10⁸ ms⁻¹, 6.2 mT
4.8 × 10⁸ ms⁻¹, 0.62 mT
5.3 × 10⁷ ms⁻¹, 4.4 mT
2.3 × 10⁷ ms⁻¹, 0.44 mT ✓

Explanation

Kinetic energy = 1.50 keV = 1500 × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁶ J.

From KE = ½mv²: v = √(2KE ÷ m) = √(4.8 × 10⁻¹⁶ ÷ 9.11 × 10⁻³¹) ≈ 2.3 × 10⁷ ms⁻¹.

In a magnetic field the force supplies the centripetal force: qvB = mv²/r, so B = mv ÷ (qr).

B = (9.11 × 10⁻³¹ × 2.3 × 10⁷) ÷ (1.6 × 10⁻¹⁹ × 0.30) ≈ 4.4 × 10⁻⁴ T = 0.44 mT. The speed is 2.3 × 10⁷ ms⁻¹.

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