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 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:
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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.
Updated On: Jun 26, 2026
- \(5\times10^5\)
- \(1\times10^6\)
- \(\pi\times10^5\)
- \(3\times10^6\)
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The Correct Option is B
Solution and Explanation
Step 1: Magnetic field inside the solenoid.
The magnetic field inside a long solenoid is \[ B=\mu_0 nI \] Here, \[ \mu_0=4\pi\times10^{-7}\ \text{T m A}^{-1} \] \[ n=5000\ \text{turns m}^{-1} \] \[ I=5\ \text{A} \] Therefore, \[ B=(4\pi\times10^{-7})(5000)(5) \] \[ B=100000\pi\times10^{-7} \] \[ B=\pi\times10^{-2}\ \text{T} \]
Step 2: Find the frequency of circular motion.
The frequency of revolution of a charged particle in a magnetic field is \[ f=\frac{qB}{2\pi m} \] Substituting the values, \[ f=\frac{(1\times10^{-16})(\pi\times10^{-2})}{2\pi(1\times10^{-27})} \] \[ f=\frac{1\times10^{-18}}{2\times10^{-27}} \] \[ f=\frac{1}{2}\times10^9 \] \[ f=5\times10^8\ \text{s}^{-1} \]
Step 3: Find the velocity component along the solenoid axis.
The particle velocity makes an angle \(60^\circ\) with the axis of the solenoid.
So, the axial component of velocity is \[ v_{\parallel}=v\cos60^\circ \] \[ v_{\parallel}=1000\times\frac{1}{2} \] \[ v_{\parallel}=500\ \text{m s}^{-1} \]
Step 4: Find the time spent inside the solenoid.
Taking the solenoid length as \[ L=1\ \text{m} \] The time taken to emerge from the opposite end is \[ t=\frac{L}{v_{\parallel}} \] \[ t=\frac{1}{500} \] \[ t=2\times10^{-3}\ \text{s} \]
Step 5: Find the number of revolutions.
The number of revolutions is \[ N=ft \] \[ N=(5\times10^8)(2\times10^{-3}) \] \[ N=10\times10^5 \] \[ N=1\times10^6 \]
Step 6: Final conclusion.
Therefore, the number of revolutions made by the particle is \[ \boxed{1\times10^6} \]
The magnetic field inside a long solenoid is \[ B=\mu_0 nI \] Here, \[ \mu_0=4\pi\times10^{-7}\ \text{T m A}^{-1} \] \[ n=5000\ \text{turns m}^{-1} \] \[ I=5\ \text{A} \] Therefore, \[ B=(4\pi\times10^{-7})(5000)(5) \] \[ B=100000\pi\times10^{-7} \] \[ B=\pi\times10^{-2}\ \text{T} \]
Step 2: Find the frequency of circular motion.
The frequency of revolution of a charged particle in a magnetic field is \[ f=\frac{qB}{2\pi m} \] Substituting the values, \[ f=\frac{(1\times10^{-16})(\pi\times10^{-2})}{2\pi(1\times10^{-27})} \] \[ f=\frac{1\times10^{-18}}{2\times10^{-27}} \] \[ f=\frac{1}{2}\times10^9 \] \[ f=5\times10^8\ \text{s}^{-1} \]
Step 3: Find the velocity component along the solenoid axis.
The particle velocity makes an angle \(60^\circ\) with the axis of the solenoid.
So, the axial component of velocity is \[ v_{\parallel}=v\cos60^\circ \] \[ v_{\parallel}=1000\times\frac{1}{2} \] \[ v_{\parallel}=500\ \text{m s}^{-1} \]
Step 4: Find the time spent inside the solenoid.
Taking the solenoid length as \[ L=1\ \text{m} \] The time taken to emerge from the opposite end is \[ t=\frac{L}{v_{\parallel}} \] \[ t=\frac{1}{500} \] \[ t=2\times10^{-3}\ \text{s} \]
Step 5: Find the number of revolutions.
The number of revolutions is \[ N=ft \] \[ N=(5\times10^8)(2\times10^{-3}) \] \[ N=10\times10^5 \] \[ N=1\times10^6 \]
Step 6: Final conclusion.
Therefore, the number of revolutions made by the particle is \[ \boxed{1\times10^6} \]
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