An electron follows a helical path in a uniform magnetic field of magnitude 0.447 T. The...

An electron follows a helical path in a uniform magnetic field of magnitude 0.447 T. The pitch of the path is 4.88 μm, and the magnitude of the magnetic force on the electron is 1.64 × 10-15N. What is the electron's speed?

ASAP

Expert Solution

w is the angular velocity, T is the time period of the helical path, q is the charge, m is the mass of the charge particle. Because the magnetic force F_B supplies the centripetal force F_c, we have

The parallel motion determines the pitch p of the helix, which is the distance between adjacent turns. This distance equals the parallel component of the velocity times the period:

Adding (1) and (2), you can obtain

Here, F_b = 1.64 x 10^-15 N, q = 1.6 x 10^-19 C, m = 9.1 x 10^-31 Kg, B = 0.447 T, p = pitch = 4.88 x 10^-6 m. Putting all the values, you can obtain

genius_generous answered 7 months ago

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