An ion with a charge of +3.25 *10⁻¹⁹ C is in region where a uniform electric field of 5 * 10^4.V/m is perpendicular to a uniform magnetic field of 0.8 T. If its acceleration is zero then its speed must be: A) 0 B) 1.6 *10⁴ m/s C) 4.0* 10⁵ m/s D) 6.3 *10⁵ m/s E) any value but 0

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Quizplus · Accepted Answer

The magnetic force on the ion is given by:$$F_m = qvB$$where:* q is the charge of the ion* v is the speed of the ion* B is the magnetic field strengthThe electric force on the ion is given by:$$F_e = qE$$where:* q is the charge of the ion* E is the electric field strengthIf the ion is in equilibrium, then the magnetic force and the electric force must be equal in magnitude and opposite in direction. Therefore:$$F_m = F_e$$$$qvB = qE$$$$v = \frac{E}{B}$$Substituting the given values into this equation, we get:$$v = \frac{5 	imes 10^4 	ext{ V/m}}{0.8 	ext{ T}} = 6.3 	imes 10^5 	ext{ m/s}$$Therefore, the speed of the ion must be 6.3 × 10^5 m/s.

An Ion with a Charge of +3