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 enters a uniform magnetic field with a velocity that is oblique to the field, its trajectory is a helix. The component of velocity perpendicular to the field causes circular motion, while the parallel component causes linear translation along the field lines (pitch).
Step 2: Key Formula or Approach:
1. Time period of one revolution: $T = \frac{2\pi m}{qB}$.
2. Pitch (distance along B-field per revolution): $P = v_{\parallel} \times T$.
3. Total distance for $N$ revolutions: $D = N \times P$.
Step 3: Detailed Explanation:
Given values:
Mass $m = 5\text{ mg} = 5 \times 10^{-6}\text{ kg}$.
Charge $q = 5\pi \times 10^{-6}\text{ C}$.
Magnetic field $\vec{B} = 0.1\hat{k}\text{ T}$.
Velocity $\vec{v} = (0.03\hat{i} + 0.02\hat{k})\text{ m/s}$.
The magnetic field is along the $z$-axis ($\hat{k}$).
The velocity component parallel to the magnetic field is $v_{\parallel} = 0.02\text{ m/s}$.
First, find the time period $T$ of revolution:
$T = \frac{2\pi m}{qB} = \frac{2\pi \times 5 \times 10^{-6}}{5\pi \times 10^{-6} \times 0.1}$.
Cancel out the common terms ($5\pi \times 10^{-6}$ in numerator and denominator):
$T = \frac{2}{0.1} = 20\text{ seconds}$.
The distance moved along the $\hat{k}$ direction in one revolution (Pitch) is:
$P = v_{\parallel} \times T = 0.02 \times 20 = 0.4\text{ m}$.
The total distance moved after 5 revolutions is:
$D = 5 \times P = 5 \times 0.4 = 2\text{ m}$.
The problem states this distance is $\alpha$ meters, so $\alpha = 2$.
Step 4: Final Answer:
The value of $\alpha$ is 2.