Question

A 5 mg particle carrying a charge of \(5\pi \times 10^{-6}\) C is moving with velocity of \((3\hat{i} + 2\hat{k}) \times 10^{-2}\) m/s in a region having magnetic field \(\vec{B} = 0.1 \hat{k}\) Wb/m². It moves a distance of \(a\) meter along \(\hat{k}\) when it completes 5 revolutions. The value of \(a\) is ________.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
When a charged particle has a velocity component parallel (\(v_{\parallel}\)) and perpendicular (\(v_{\perp}\)) to a uniform magnetic field, it follows a helical path. The time taken for one revolution is determined by the perpendicular component, but the distance moved along the field lines depends on the parallel component.
Step 2: Key Formula or Approach:
1. Time period of one revolution: \(T = \frac{2\pi m}{qB}\)
2. Pitch (distance in 1 revolution): \(p = v_{\parallel} \times T\)
3. Total distance: \(a = n \times p\), where \(n\) is the number of revolutions.
Step 3: Detailed Explanation:
Given: \(m = 5 \text{ mg} = 5 \times 10^{-6}\) kg, \(q = 5\pi \times 10^{-6}\) C, \(B = 0.1\) T. Velocity \(\vec{v} = (3\hat{i} + 2\hat{k}) \times 10^{-2}\). Here, \(v_{\parallel} = 2 \times 10^{-2}\) m/s (component along \(\vec{B}\)). Calculate Time Period \(T\): \[ T = \frac{2\pi \times 5 \times 10^{-6}}{(5\pi \times 10^{-6}) \times 0.1} = \frac{10\pi}{0.5\pi} = 20 \text{ s} \] Distance for 5 revolutions: \[ a = 5 \times (v_{\parallel} \times T) \] \[ a = 5 \times (2 \times 10^{-2}) \times 20 \] \[ a = 5 \times 0.4 = 2 \text{ m} \]
Step 4: Final Answer:
The value of \(a\) is 2.