A charged particle (m = 5.0 g,q = -70 $\mu$C)moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s² downward,the electric field is 700 N/C upward,and the magnetic field is perpendicular to the velocity of the particle.What is the magnitude of the magnetic field in this region?

Question

Quizplus · Accepted Answer

The magnetic force on the particle is equal to the gravitational force acting on it. Therefore, the magnitude of the magnetic field is:$$B = \frac{mg}{qv}$$$$B = \frac{(5.0 	imes 10^{-3} 	ext{ kg})(9.8 	ext{ m/s}^2)}{(1.0 	imes 10^{-6} 	ext{ C})(30 	imes 10^3 	ext{ m/s})}$$$$B = 0.047 	ext{ T} = 47 	ext{ mT}$$

A Charged Particle (M = 5 μ\muμ C)moves Horizontally at a Constant Speed of 30 Km/s in C)moves

A Charged Particle (M = 5 $\mu$ C)moves Horizontally at a Constant Speed of 30 Km/s in C)moves