Helical Motion of a Proton

Helical Motion of a Proton

Question: A proton enters a magnetic field at \(60^\circ\). The radius of its helical path is:

(a) 1.3 cm

(b) 2.6 cm

(c) 4.5 cm

(d) 13 cm

Solution

Step 1: Perpendicular Velocity Component
\[ v_\perp = v \sin\theta = 5 \times 10^6 \cdot \sin 60^\circ = 4.33 \times 10^6 \, \text{m/s} \]

Step 2: Radius Formula
\[ r = \frac{m v_\perp}{qB} = \frac{(1.67 \times 10^{-27})(4.33 \times 10^6)}{(1.6 \times 10^{-19})(0.2)} \approx 0.026 \, \text{m} = 2.6 \, \text{cm} \]

Answer: Option (b) \(2.6 \, \text{cm}\) is correct.