   Chapter 19, Problem 18P

Chapter
Section
Textbook Problem

Electrons in Earth’s upper atmosphere have typical speeds near 6.00 × 105 m/s. (a) Calculate the magnitude of Earth’s magnetic field if an electron’s velocity is perpendicular to the magnetic field and its circular path has a radius of 7.00 × 10−2 m. (b) Calculate the number of times per second that an electron circles around a magnetic field line.

a)

To determine
The magnitude of earth’s magnetic field.

Explanation

Given info: The speed of the electron is 6.00×105ms-1 . The earth’s magnetic field is perpendicular to the electrons velocity. The radius of its circular path is 7.00×102m .

Explanation:

The centripetal force is supplied by the magnetic field.

Hence,

qvB=mv2r

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

Re-arranging for the magnetic field,

B=mvqr

Substitute 9.11×1031kg for m , 6.00×105ms-1 for v , 1.60×1019C for q and 7

b)

To determine
The number of times per second that an electron circles around a magnetic field line.

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