Question

A particle of mass \(1\times10^{-27}\ \text{kg}\) and charge \(1\times10^{-16}\ \text{C}\) enters the uniform magnetic field inside a solenoid at speed \(1000\ \text{m s}^{-1}\). The velocity vector makes an angle \(60^\circ\) with the axis of the solenoid. The solenoid has \(5000\) turns along its length and carries current \(5\ \text{A}\). The number of revolutions the particle makes along the helical path within the solenoid by the time it emerges from the solenoid's opposite end is:

• A. \(5\times10^5\)
• B. \(1\times10^6\)
• C. \(\pi\times10^5\)
• D. \(3\times10^6\)

Hint:

In a uniform magnetic field, the component of velocity perpendicular to the field causes circular motion, while the component parallel to the field causes forward motion. Together, they produce a helical path.

Answer 1

The Correct Option is B

Step 1: Calculate the magnetic field inside the solenoid.
The magnetic field inside a long solenoid is: \[ B = \mu_0 n I \] where $n = 5000\,\text{turns m}^{-1}$ and $I = 5\,\text{A}$: \[ B = (4\pi \times 10^{-7})(5000)(5) = \pi \times 10^{-2}\,\text{T} \]
Step 2: Find the cyclotron frequency of the particle.
A charged particle moving perpendicular to a magnetic field undergoes circular motion. The cyclotron frequency (revolutions per second) is: \[ f = \frac{qB}{2\pi m} \] Substituting $q = 1 \times 10^{-16}\,\text{C}$, $B = \pi \times 10^{-2}\,\text{T}$, $m = 1 \times 10^{-27}\,\text{kg}$: \[ f = \frac{(10^{-16})(\pi \times 10^{-2})}{2\pi (10^{-27})} = \frac{\pi \times 10^{-18}}{2\pi \times 10^{-27}} = \frac{10^{-18}}{2 \times 10^{-27}} = 5 \times 10^{8}\,\text{Hz} \]
Step 3: Find the axial (parallel) component of velocity.
The velocity makes $60^\circ$ with the solenoid axis. The component along the axis is responsible for the axial translation (forming the helix): \[ v_\parallel = v \cos 60^\circ = 1000 \times \frac{1}{2} = 500\,\text{m s}^{-1} \]
Step 4: Find the pitch of the helical path.
The pitch is the axial distance traveled per revolution: \[ p = \frac{v_\parallel}{f} = \frac{500}{5 \times 10^{8}} = 10^{-6}\,\text{m} \]
Step 5: Find the length of the solenoid and number of revolutions.
For a solenoid with $n = 5000\,\text{turns m}^{-1}$ and total turns $N = 5000$ (implying $L = 1\,\text{m}$): \[ \text{Number of revolutions} = \frac{L}{p} = \frac{1}{10^{-6}} = 10^6 \]
Step 6: State the final answer.
The number of revolutions made by the particle along the helical path is: \[ \boxed{1 \times 10^6} \]