Question

Achargedparticle(m=5.0g,q=−70μC)moveshorizontallyataconstantspeedof30km/sina 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?

Answer 1

Magnetic field, 0.046 T

Step-by-step explanation:

Given that,

Mass of the charged particle, m = 5 g

Charge,
`q=-70\ \mu C=-70* 10^(-6)\ C`

Electric field, E = 700 N/C

Speed of the charge particle, v = 30 km/s = 30000 m/s

To find,

The magnitude of the magnetic field in this region.

Solution,

The charged particle is moving horizontally with a constant speed, the gravitational force is balanced by the magnetic force and the electric force.
`mg=qvB+qE`

`mg-qE=qvB`

`B=(mg-qE)/(qv)`

`B=(5* 10^(-3)* 9.8-(-70* 10^(-6))* 700)/(70* 10^(-6)* 30000)`

B = 0.046 T

Therefore, the magnitude of the magnetic field in this region is 0.046 T. Hence, this is the required solution.