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Find the radius of an orbit of an electron moving at a rate of 2.0 x 10² m/s in a uniform magnetic field of 1.20 x 10 ̄³ T.

Correct answer: C. 9.48 x 10 ̄² m

  • A. 3.15 x 10 ̄² m
  • B. 4.25 x 10 ̄² m
  • C. 9.48 x 10 ̄² m
  • D. 17.77 x 10 -2 m
  • E. 19.91 x 10 ̄² m

Explanation

The formula for the radius isr = mv/qB= (9.11 x 10-31) (2 x 102) / (1.6 x 10-19) (1.2 x 10-3)=9.48 x 10-7mHence the correct option is C.

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About Electromagnetism

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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