Question

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?

Answer 1

B=0.047mT

Step-by-step explanation:

The magnetic field in this region is given by the followin equation
`F_g=F_b+F_e`

Where
`F_g = mg, F_b=qvB, F_e=qE`

Replacing all of this values,

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

We have,

`mg=qvB+qE\\(0.005)(9.8) = (-70*10^(-6))(30*10^3)(B)-(70*10^(-6))(700)\\B=0.047T\\B=0.047mT`