Question

An alpha particle (mass \(6.4 \times 10^{-27} \text{ kg}\) and charge \(3.2 \times 10^{-19} \text{ C}\)) having \(8.0 \text{ MeV}\) energy, enters a region of a uniform magnetic field of \(0.5 \text{ T}\). 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:

Remember that \(r = \frac{\sqrt{2mK}}{qB}\) is the quickest way to find the radius when dealing with energy. Ensure kinetic energy is always scaled out of MeV directly into Joules by multiplying by \(1.6 \times 10^{-13}\).

Answer 1

Concept: When a charged particle of charge \(q\) and mass \(m\) moves with a velocity \(v\) inside a uniform magnetic field \(B\), it experiences a magnetic Lorentz force given by \(\vec{F} = q(\vec{v} \times \vec{B})\). If the velocity vector is perpendicular to the magnetic field line vector (\(\phi = 90^{\circ}\)), the force acts as a centripetal force: \[ \frac{m v^2}{r} = q v B \quad \Rightarrow \quad r = \frac{m v}{q B} \] Since momentum \(p = m v\) can be written in terms of kinetic energy \(K\) as \(p = \sqrt{2 m K}\), the radius can be directly computed using: \[ r = \frac{\sqrt{2 m K}}{q B} \]

Step 1: Listing variables and changing units into S.I. system.
The values provided for the incoming alpha particle are: Mass, m &= 6.4 \times 10^{-27} kg
Charge, q &= 3.2 \times 10^{-19} C
Magnetic Field, B &= 0.5 T
Kinetic Energy, K &= 8.0 MeV = 8.0 \times 10^6 eV Converting electron-volts (eV) into standard Joules (J): \[ K = 8.0 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 12.8 \times 10^{-13} \text{ J} = 1.28 \times 10^{-12} \text{ J} \]

Step 2: Computing the radius of the circular path.
Using our formula: \[ r = \frac{\sqrt{2 \times (6.4 \times 10^{-27} \text{ kg}) \times (1.28 \times 10^{-12} \text{ J})}}{(3.2 \times 10^{-19} \text{ C}) \times 0.5 \text{ T}} \] First, evaluate the expression inside the square root in the numerator: \[ \text{Inside Value} = 2 \times 6.4 \times 1.28 \times 10^{-27} \times 10^{-12} = 16.384 \times 10^{-39} = 1.6384 \times 10^{-38} \] Taking the square root of this value: \[ \sqrt{1.6384 \times 10^{-38}} = 1.28 \times 10^{-19} \text{ kg}\cdot\text{m/s} \] Now evaluate the denominator: \[ \text{Denominator} = 3.2 \times 10^{-19} \times 0.5 = 1.6 \times 10^{-19} \text{ C}\cdot\text{T} \] Substitute these simplified segments back into the calculation for radius \(r\): \[ r = \frac{1.28 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{1.28}{1.6} = 0.8 \text{ m} \] Correction based on exact mass value representations: Let us crosscheck with alternative representations: \[ r = \frac{\sqrt{2 \times 6.4 \times 10^{-27} \times 8 \times 10^6 \times 1.6 \times 10^{-19}}}{3.2 \times 10^{-19} \times 0.5} \] \[ r = \frac{\sqrt{163.84 \times 10^{-40}}}{1.6 \times 10^{-19}} = \frac{12.8 \times 10^{-20}}{1.6 \times 10^{-19}} = 0.8 \text{ m} \] *(Note: If standard values for alpha particle are taken where \(m = 4 \text{ amu}\), variations can exist; following the given question values yields precisely \(0.80\text{ m}\). Let's evaluate closer to option structures if another configuration was implied).*

Step 3: Analyzing special motion conditions.

• (i) Condition for helical path: The particle will describe a helical path if its velocity vector \(\vec{v}\) makes an oblique angle \(\phi\) with the magnetic field lines such that \(\phi \neq 0^{\circ}, 90^{\circ}, \text{ or } 180^{\circ}\). The component \(v \sin\phi\) provides circular motion while \(v \cos\phi\) drives linear translation.
• (ii) Condition for going straight and undeviated: The particle travels completely straight and undeviated if the magnetic Lorentz force acting on it is zero. This happens when the velocity vector is parallel or anti-parallel to the field line lines (\(\phi = 0^{\circ}\) or \(\phi = 180^{\circ}\)), meaning \(\vec{v} \times \vec{B} = 0\). Alternatively, this can occur if a balancing electric field \(\vec{E}\) is introduced such that the net Lorentz force is zero (\(\vec{E} = - \vec{v} \times \vec{B}\)).