   Chapter 19, Problem 66AP

Chapter
Section
Textbook Problem

An electron moves in a circular path perpendicular to a constant magnetic field of magnitude 1.00 mT. The angular momentum of the electron about the center of the circle is 4.00 × 10−25 kg · m2/s. Determine (a) the radius of the circular path and (b) the speed of the electron.

a)

To determine
The radius of the circular path of the electron.

Explanation

Given info: The magnitude of the constant magnetic field is 1.00mT . The angular momentum of the electron about the center of the circle is 4.00×1025kgm2s-1 .

Explanation:

The angular momentum of the electron is,

L=mvr       (1)

• m is the mass of the electron
• v is the velocity of the electron
• r is the radius of the circular path of the electron

Since the magnetic field supplies the centripetal acceleration,

qvB=mv2r

• B is the magnetic field present
• v is the velocity
• q is the charge of the particle
• r is the radius of the circular path
• m is the mass of the particle

Re-arranging,

mv=Brq       (2)

Combining (1) and (2),

L=Bqr2

The radius of the path will be,

r=LBq

Substitute 1.00mT for B , 4.00×1025kgm2s-1 for L , 1

b)

To determine
The speed of the electron.

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