Final answer:
The magnitude of the force exerted on an electron moving at 3.0×10⁶ m/s in a 3.0×10⁻⁴ T magnetic field, assuming a perpendicular angle, can be calculated to be 1.44×10⁻ⁱ⁶ N using the formula F = qvBsin(θ).
Step-by-step explanation:
To calculate the magnitude of the force exerted on an electron moving through a magnetic field, we use the formula F = qvBsin(θ), where F is the force, q is the charge of the electron (approximately -1.6 × 10-19 C), v is the velocity of the electron, B is the magnetic field, and θ is the angle between the velocity and the magnetic field.
In the given case, the magnetic field B is 3.0 × 10-4 T, and the velocity v of the electron is 3.0 × 106 m/s. Assuming the electron moves perpendicular to the magnetic field (θ = 90°), the magnitude of the force can be calculated.
Applying the formula:
F = (1.6 × 10-19 C)(3.0 × 106 m/s)(3.0 × 10-4 T)sin(90°)
F = 1.6 × 10-19 × 3.0 × 106 × 3.0 × 10-4
F = 1.44 × 10-16 N.
Therefore, the magnitude of the force exerted on the electron is 1.44 × 10-16 N.