Question

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?

Answer 1

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

Step-by-step 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.

Answer 2

The magnitude of the force, B = 5 Tesla, Up (North) direction

Step-by-step explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 ×
`10^(-5)`C

F = 750 × -7 ×
`10^(-5)`

F = 0.0525

But F = qvB; B =
`(F)/(qv)`

where B is the magnetic field

= 0.0525 ÷ ( -7 ×
`10^(-5)` × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).