Find the radius of an orbit of an electron moving at a rate of 2.0 x 10^7 m/s in a uniform magnetic field of 1.20 x 10^-3 T.
Correct answer: C. 9.93 m
- A. 3.15 m
- B. 4.25 m
- C. 9.93 m
- D. 17.77 m
- E. 19.91 m
Explanation
r = (m * v) / (q * B)Given:m = 9.109 x 10^-31 kgv = 2.0 x 10^7 m/sq = -1.6 x 10^-19 CB = 1.20 x 10^-3 TNow, let's calculate the radius (r):r = (9.109 x 10^-31 kg * 2.0 x 10^7 m/s) / (1.6 x 10^-19 C * 1.20 x 10^-3 T)r ≈ 9.93 mThe correct option is C. 9.93 m.
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Magnetic flux density and magnetic flux describe the strength of a magnetic field and the field passing through a surface. A charged particle moving through a magnetic field experiences a force perpendicular to its velocity and may follow circular or helical motion, depending on the angle between velocity and field.
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