A charged particle (mass = 4.0 μg, charge = 5.0 μC) moves in a region where the only force on it is magnetic. What is the magnitude of the acceleration of the particle at a point where the speed of the particle is 5.0 km/s, the magnitude of the magnetic field is 8.0 mT, and the angle between the direction of the magnetic field and the velocity of the particle is 60°? A) 39 km/s² B) 43 km/s² C) 48 km/s² D) 52 km/s² E) 25 km/s²

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Quizplus · Accepted Answer

The force on a charged particle moving in a magnetic field is given by the Lorentz force equation: $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity of the particle, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field direction. The acceleration $a$ can be found using Newton's second law, $F = ma$, where $m$ is the mass of the particle. Substituting the given values (converting units where necessary: $4.0 \mu g = 4.0 \times 10^{-9} kg$, $5.0 \mu C = 5.0 \times 10^{-6} C$, $5.0 km/s = 5000 m/s$, $8.0 mT = 8.0 \times 10^{-3} T$), and using $\sin(60°) = \sqrt{3}/2$, we get:$$F = (5.0 \times 10^{-6} C)(5000 m/s)(8.0 \times 10^{-3} T)\sin(60°) = (5.0 \times 10^{-6})(5000)(8.0 \times 10^{-3})(\sqrt{3}/2) N$$$$F = 1.04 \times 10^{-1} \sqrt{3} N$$$$a = \frac{F}{m} = \frac{1.04 \times 10^{-1} \sqrt{3} N}{4.0 \times 10^{-9} kg}$$$$a = 26 \times 10^{3} \sqrt{3} m/s^2$$$$a \approx 43 \times 10^{3} m/s^2 = 43 km/s^2$$Therefore, the correct answer is B, 43 km/s^2.

A Charged Particle (Mass = 4