A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 km/s in a region where the free fall gravitational acceleration is 9.81 m/s2 downward, the electric field is 750 N/C upward and the magnetic field is perpendicular to the velocity of the particle. What is the magnitude and direction of the magnetic field in this region?

The magnitude  of the magnetic field is 0.025 T in upward direction.

Explanation:

Given;

charge of the particle, q = -70.0μC

mass of the particle, m = 5 g = 0.005 kg

velocity of the particle, v = 30 km/s

acceleration due to gravity, g = 9.81 m/s²

electric field strength, E = 750 N/C

electric force on the particle = Eq

magnetic force on the particle = qvBsinθ

since the magnetic field is perpendicular to the velocity of the particle, θ = 90

Eq =  qvB

E = vB

Where;

E is the electric field

B is the magnetic field

v is the velocity of the particle

B = E/v

B = 750/30000

B = 0.025 T upward.

Therefore, the magnitude  of the magnetic field is 0.025 T in upward direction.

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