Protons travel right through crossed fields: E = 8.00 × 10⁵ V m⁻¹ upward and B = 0.200 T out of the page. The undeflected beam then enters a region containing only the same magnetic field. Determine the selected speed, force directions in the selector, circular radius, orbital period, and whether doubling speed changes the period. Use proton mass 1.67 × 10⁻²⁷ kg and charge 1.602 × 10⁻¹⁹ C.

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v = 4.00 × 10⁶ m s⁻¹; electric force upward and magnetic force downward; r = 0.209 m; T = 3.28 × 10⁻⁷ s. Doubling speed doubles radius but not period. For a positive proton, electric force follows E and points upward. With velocity right and B out of the page, v × B points downward, so magnetic force can cancel it. Undeflected motion requires qE = qvB. Charge cancels, giving v = E/B = (8.00 × 10⁵)/0.200 = 4.00 × 10⁶ m s⁻¹. In the magnetic-only region, qvB supplies centripetal force: qvB = mv²/r. Thus r = mv/(qB) = 0.209 m. The period is circumference divided by speed: T = 2πr/v = 2πm/(qB) = 3.28 × 10⁻⁷ s. Doubling v doubles r because magnetic force scales with v while required centripetal force scales with v²/r. The ratio r/v and therefore the non-relativistic period remain unchanged.

Select and bend a proton beam

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