Final answer:
The speed of electrons bent by a uniform magnetic field in a circle of radius 2.0 cm can be calculated using the Lorentz force equation; it comes out to approximately 2.9×10· m/s.
Step-by-step explanation:
To calculate the speed of electrons bent in a circle of radius 2.0 cm by a uniform magnetic field with B=4.5×10⁻³ T, we use the principle that the magnetic force acting on the electron provides the centripetal force necessary to keep it moving in a circular path. The force exerted by a magnetic field on a moving charge is given by the Lorentz force equation, F = qvBsin(θ), where q is the charge of the electron, v is the velocity of the electron, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field vectors.
Since the electron is moving in a circle perpendicular to the magnetic field, θ is 90 degrees and sin(θ) = 1. Assuming that the mass of an electron is m = 9.11×10⁻³± kg, the centripetal force necessary to maintain circular motion is F = mv²/r. Therefore, the magnetic force serves as the centripetal force, and we have qvB = mv²/r. Simplifying this, we can solve for the speed of the electron:
v = qBr/m
Where q = charge of the electron (1.60×10⁻ C), r = radius of the circle, m = mass of the electron. Plugging in the values:
v = (1.60×10⁻ C)(4.5×10⁻³ T)(0.02 m) / (9.11×10⁻³± kg)
After calculation, the speed v is approximately 2.9×10· m/s, which matches answer choice A.