Achargedparticle(m=5.0g, q=−70μc)30km/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?

Magnetic field, 0.046 T

Explanation:

Given that,

Mass of the charged particle, m = 5 g

Charge,

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.

B = 0.046 T

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

answer: a. same unit

explanation:

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr