A charged particle (q = 2 × 10⁻⁸ C, m = 4 × 10⁻⁶ kg) moves at 4 × 10³ m/s in a 0.1 T magnetic field at 30°. The radius of its helical path is:
Correct answer: C. 0.08 m
- A. 0.02 m
- B. 0.04 m
- C. 0.08 m
- D. 0.12 m
Explanation
Radius r = mv /(qB), where v = v sinθ = 4 × 10³ × sin 30° = 2 × 10³ m/s. Substituting m = 4 × 10⁻⁶ kg, q = 2 × 10⁻⁸ C, B = 0.1 T. r = (4 × 10⁻⁶ × 2 × 10³)/(2 × 10⁻⁸ × 0.1) = 0.08 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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