Question

An alpha particle (mass $6.4 \times 10^{-27}$ kg and charge $3.2 \times 10^{-19}$ C) having 8.0 MeV energy, enters a region of a uniform magnetic field of 0.5 T. If the field is directed perpendicular to the velocity of the particle, find the radius of the circular path described by the particle. Mention the condition under which the particle in this region (i) describes a helical path, and (ii) goes straight undeviated.

Hint:

Notice how $2 \times 6.4 = 12.8$, which perfectly matched the $12.8$ from the kinetic energy. Examiners often design the numbers in these square roots to form perfect squares. Always group terms carefully before multiplying blindly!

Answer 1

Concept:
• When a charged particle enters a uniform magnetic field perpendicularly, it experiences a maximum Lorentz force that acts entirely as a centripetal force, causing it to move in a perfectly circular path.
• The radius of this circular path is derived by equating the magnetic force to the centripetal force: $q v B = \frac{m v^2}{r} \implies r = \frac{m v}{q B} = \frac{p}{q B}$.
• The momentum $p$ can be related to the kinetic energy $K$ via the relation $p = \sqrt{2 m K}$.

Step 1: Identify and convert the given parameters
Mass of alpha particle, $m = 6.4 \times 10^{-27} \text{ kg}$.
Charge of alpha particle, $q = 3.2 \times 10^{-19} \text{ C}$.
Magnetic field strength, $B = 0.5 \text{ T}$.
Kinetic Energy, $K = 8.0 \text{ MeV}$.
First, convert the kinetic energy from MeV to Joules (Standard SI unit):
\[ K = 8.0 \times 10^6 \text{ eV} \]
\[ K = 8.0 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} \]
\[ K = 12.8 \times 10^{-13} \text{ J} \]

Step 2: Calculate the momentum of the alpha particle
Use the energy-momentum relation:
\[ p = \sqrt{2 m K} \]
\[ p = \sqrt{2 \times (6.4 \times 10^{-27}) \times (12.8 \times 10^{-13})} \]
\[ p = \sqrt{12.8 \times 10^{-27} \times 12.8 \times 10^{-13}} \]
\[ p = \sqrt{(12.8)^2 \times 10^{-40}} \]
\[ p = 12.8 \times 10^{-20} \text{ kg m/s} \]

Step 3: Calculate the radius of the circular path
Substitute momentum into the radius formula:
\[ r = \frac{p}{q B} \]
\[ r = \frac{12.8 \times 10^{-20}}{(3.2 \times 10^{-19}) \times 0.5} \]
\[ r = \frac{12.8 \times 10^{-20}}{1.6 \times 10^{-19}} \]
\[ r = 8 \times 10^{-1} \text{ m} \]
\[ r = 0.8 \text{ m} \]

Step 4: Provide conditions for helical and undeviated paths
(i) Helical Path: The particle will describe a helical path if its initial velocity vector makes an oblique angle $\theta$ (where $0^\circ < \theta < 90^\circ$ or $90^\circ < \theta < 180^\circ$) with the direction of the uniform magnetic field. The perpendicular velocity component provides the circular motion, while the parallel component provides the linear translation.
(ii) Straight Undeviated Path: The particle will pass straight through undeviated if it is injected exactly parallel or exactly anti-parallel to the magnetic field lines ($\theta = 0^\circ$ or $\theta = 180^\circ$). Under these conditions, the magnetic force $F_m = q v B \sin\theta$ becomes entirely zero.

Step 5: Conclusion
The radius of the circular path is $0.8 \text{ m}$. A helical path requires an oblique entry angle, and an undeviated straight path requires parallel or anti-parallel entry relative to the magnetic field.