JEE Challenger
More from Moving Charges and Magnetism
PhyMedAdv 2021 • Paper 1

Ratio of Circular Orbits for Alpha Particle and Sulfur Ion in Magnetic Field

An α\alpha-particle (having a mass of 4 amu4\text{ amu}) and a singly charged sulfur ion (having a mass of 32 amu32\text{ amu}) start from rest. Both are accelerated across a potential difference VV and enter a uniform magnetic field directed perpendicular to their velocities. In this magnetic field, the α\alpha-particle and the sulfur ion follow circular paths of radii rαr_{\alpha} and rSr_S, respectively. Determine the ratio (rS/rα)(r_S/r_{\alpha}).

Official Numerical Answer4

Topics & Concepts

Moving Charges and MagnetismMagnetic ForceElectric Potential

Step-by-Step Solution

When a charged particle of mass mm and charge qq is accelerated from rest through a potential difference VV, it acquires a kinetic energy K=qVK = qV. Upon entering a perpendicular uniform magnetic field BB, it moves in a circular path of radius:

r=mvqB=2mKqB=1B2mVqr = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}

This indicates that the orbital radius is proportional to m/q\sqrt{m/q}.

For the α\alpha-particle: mα=4 amu,qα=2em_\alpha = 4\text{ amu}, \quad q_\alpha = 2e

For the singly charged sulfur ion: mS=32 amu,qS=em_S = 32\text{ amu}, \quad q_S = e

Taking the ratio of their radii:

rSrα=(mSmα)(qαqS)=(324)(2ee)=8×2=4\frac{r_S}{r_\alpha} = \sqrt{\left(\frac{m_S}{m_\alpha}\right) \left(\frac{q_\alpha}{q_S}\right)} = \sqrt{\left(\frac{32}{4}\right) \left(\frac{2e}{e}\right)} = \sqrt{8 \times 2} = 4

Thus, the ratio (rS/rα)(r_S/r_\alpha) is 44.

PreviousNext