A charged particle (q = +2 μC, m = 2 × 10⁻⁶ kg) moves at 5 × 10⁴ m/s perpendicular to a magnetic field of 0.4 T. The radius of its circular path is:
Correct answer: B. 0.025 m
- A. 0.0125 m
- B. 0.025 m
- C. 0.05 m
- D. 0.1 m
Explanation
The radius of a charged particle's circular path in a magnetic field is given by r = mv/(qB). Substituting m = 2 × 10⁻⁶ kg, v = 5 × 10⁴ m/s, q = 2 × 10⁻⁶ C, B = 0.4T. r = (2 × 10⁻⁶ × 5 × 10⁴)/(2 × 10⁻⁶ × 0.4) = 0.025 m.
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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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