Helical Path Radius in Magnetic Field

Helical Motion in Magnetic Field

Question: A charge with \( \frac{e}{m} = 10^8 \, \text{C/kg} \) and velocity \( 3 \times 10^5 \, \text{m/s} \) enters a uniform magnetic field \( B = 0.3 \, \text{T} \) at \( 30^\circ \). The radius of curvature is:

(a) 0.01 cm

(b) 0.5 cm

(c) 1 cm

(d) 2 cm

Solution

Step 1: Identify perpendicular velocity component
\( v_\perp = v \sin\theta = 3 \times 10^5 \, \text{m/s} \times \sin 30^\circ = 1.5 \times 10^5 \, \text{m/s} \)

Step 2: Use radius formula for circular motion
\[ r = \frac{m v_\perp}{e B} = \frac{v_\perp}{(e/m) B} = \frac{1.5 \times 10^5}{10^8 \times 0.3} \]

Step 3: Calculate the radius
\[ r = \frac{1.5 \times 10^5}{3 \times 10^7} = 5 \times 10^{-3} \, \text{m} = 0.5 \, \text{cm} \]

Answer: Option (b) \( 0.5 \, \text{cm} \) is correct.