An electron moves in a region where the magnetic field is uniform and has a magnitude of 80$\mu$T.The electron follows a helical path which has a pitch of 9.0 mm and a radius of 2.0 mm.What is the speed of this electron as it moves in this region?

Question

Quizplus · Accepted Answer

The equation for the helical path followed by a charged particle in a uniform magnetic field is given by:

R = mv/qB

where R is the radius of the helical path, m is the mass of the particle, v is the velocity of the particle, q is the charge of the particle, and B is the magnetic field.

In this case, the radius R = 2.0 mm and the pitch p = 9.0 mm. We can find the angle that the path makes with the magnetic field by noting that one complete turn of the helix corresponds to a distance of the pitch, so:

2πR = pcosθ

where θ is the angle between the magnetic field and the velocity of the electron. Solving for θ, we find:

θ = cos^-1(2πR/p) = cos^-1(2π(2.0x10^-3 m)/(9.0x10^-3 m)) ≈ 86.4°

The velocity of the electron can be found from the centripetal force acting on it due to the magnetic field:

mv^2/R = qvB

Solving for v, we find:

v = qBR/m = (1.6x10^-19 C)(80 T)(2.0x10^-3 m) / (9.11x10^-31 kg) ≈ 3.54x10^7 m/s = 35 km/s

Therefore, the correct answer is D.

An Electron Moves in a Region Where the Magnetic Field μ\muμ

An Electron Moves in a Region Where the Magnetic Field $\mu$