An electron moves in a region where the magnetic field is uniform,has a magnitude of 60$\mu$T,and points in the positive x direction.At t = 0 the electron has a velocity that has an x component of 30 km/s,a y component of 40 km/s,and a z component of zero.What is the radius of the resulting helical path?
A)4.7 mm
B)18 mm
C)3.8 mm
D)2.8 mm
E)5.7 mm

Question

Quizplus · Accepted Answer

The force on the electron due to the magnetic field is given by the equation:

F = qV x B

where F is the force, q is the charge of the electron, V is its velocity and B is the magnetic field.

In this case, the force is perpendicular to both the velocity and the magnetic field, and is given by:

F = qVB

The centripetal force required to keep the electron in a circular path is:

F = m(v^2 / r),

where m is the mass of the electron, v is its velocity, and r is the radius of its path.

Setting these two equations equal to each other and solving for r gives:

r = mv / (qB)

Plugging in the given values, we have:

r = (9.11 x 10^-31 kg)(50 x 10^3 m/s) / (1.6 x 10^-19 C)(60 T) = 3.8 x 10^-3 m = 3.8 mm (rounded to two significant figures)

Therefore, the best answer is (C).

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

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