Positive singly charged ions pass undeflected through E = 2.4 × 10⁴ V/m and B = 0.080 T, then curve at radius 0.120 m in B = 0.150 T. Charge is 1.60 × 10⁻¹⁹ C. (a) Find speed and mass. (b) Give force direction for east velocity and downward B. (c) Explain magnetic work. (d) A 50 Hz supply is 170 V peak: find rms voltage and explain why unsynchronised a.c. cannot select steadily.

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Speed 3.0 × 10⁵ m/s; mass 9.60 × 10⁻²⁷ kg; force north; Vrms ≈120 V; alternating field changes selection with entry phase. Undeflected condition qE = qvB gives v = E/B = 2.4 × 10⁴/0.080 = 3.0 × 10⁵ m/s. In analyser, qvB = mv²/r, so m = qBr/v = 9.60 × 10⁻²⁷ kg. For positive charge, v × B with east velocity and downward field points north; negative charge reverses direction. Magnetic force stays perpendicular to velocity, so F·v = 0. It changes direction without speed or kinetic-energy change. Vrms = 170/√2 = 120 V. An unsynchronised a.c. electric field changes magnitude and direction during ion entry, so balance and deflection depend on phase rather than selecting one steady speed.

Velocity selector and alternating supply

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