Protons of mass 1.67 × 10⁻²⁷ kg and charge +1.60 × 10⁻¹⁹ C move right through crossed fields: electric field 36000 V/m downward and magnetic field 0.100 T into the page. Undeflected protons then enter a region containing only a 0.200 T magnetic field into the page. (a) Find selected speed. (b) Find the circular-path radius and initial deflection direction. (c) Explain whether the magnetic field changes proton kinetic energy.

Question

Accepted Answer

Speed 360000 m/s; radius 0.01879 m; initial deflection upward; kinetic energy is unchanged by the magnetic field. For a positive proton moving right in a magnetic field into the page, the magnetic force is upward. Its electric force is downward. Undeflected motion requires equal magnitudes qE = qvB, so charge cancels. The selected speed is v = E/B = 36000/0.100 = 360000 m/s. Particles of other speeds are not balanced by this particular pair of fields, even though their charges have the same sign. In the magnetic-only region, qvB = mv²/r. Therefore r = mv/(qB) = (1.67 × 10⁻²⁷ × 360000)/(1.60 × 10⁻¹⁹ × 0.200) = 0.01879 m, with initial curvature upward. The magnetic force remains perpendicular to instantaneous velocity, so it does no work on the proton. It changes the velocity direction without changing speed or kinetic energy. An electric field could do work, but it is absent from the analysing region described here.

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