contestada

A charged particle ( m = 5.0 g, q = −70 μC) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s 2 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?

Respuesta :

cjmejiab

Answer:

B=0.047mT

Explanation:

The magnetic field in this region is given by the followin equation[tex]F_g=F_b+F_e[/tex]

Where [tex]F_g = mg, F_b=qvB, F_e=qE[/tex]

Replacing all of this values,

m = 5.0 g, q = −70 μC, g= 9.8 m/s^2, E=700N/C

We have,

[tex]mg=qvB+qE\\(0.005)(9.8) = (-70*10^{-6})(30*10^3)(B)-(70*10^{-6})(700)\\B=0.047T\\B=0.047mT[/tex]

Otras preguntas