An electron (q = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg) enters a uniform magnetic field of 0.2 T at 2 × 10⁶ m/s at an angle of 60° to the field. The radius of its helical path is:
Correct answer: B. 4.92 × 10⁻⁵ m
- A. 2.84 × 10⁻⁵ m
- B. 4.92 × 10⁻⁵ m
- C. 5.68 × 10⁻⁵ m
- D. 8.52 × 10⁻⁵ m
Explanation
The radius of a helical path is r = mv/(qB), where v = v sinθ is the velocity component perpendicular to the magnetic field. For v = 2 × 10⁶ m/s, θ = 60°, sin 60° = √3/2, q = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg, B = 0.2 T. v = 2 × 10⁶ × √3/2 ≈ 1.732 × 10⁶ m/s. Thus, r = (9.1 × 10⁻³¹ × 1.732 × 10⁶)/(1.6 × 10⁻¹⁹ × 0.2) ≈ 4.92 × 10⁻⁵ 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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