AIIMS 2004 Physics Motion in Magnetic Field MCQ Question

To find the ratio of the radii of the circular paths described by a proton and an α-particle (helium nucleus) in a uniform magnetic field, we need to use the formula for the radius of the circular motion of a charged particle in a magnetic field.

Key Concepts

1. Magnetic Force and Circular Motion: When a charged particle moves in a magnetic field at an angle perpendicular to the field, it experiences a magnetic force that acts as a centripetal force, causing it to move in a circular path.

   The magnetic force $F$ on a charged particle is given by: $F = qvB$ where:

   • $q$ is the charge of the particle,
   • $v$ is the velocity of the particle,
   • $B$ is the magnetic field strength.

2. Centripetal Force: The centripetal force required to keep a particle moving in a circular path is expressed as: $F = \frac{mv^2}{r}$ where:

   • $m$ is the mass of the particle,
   • $r$ is the radius of the circular path.

Combining the Equations

Setting the magnetic force equal to the centripetal force gives us: $qvB = \frac{mv^2}{r}$

Rearranging this to solve for the radius $r$ gives: $r = \frac{mv}{qB}$

Finding the Radii for Proton and α-Particle

1. Proton:

   • Charge $q_p = e$ (elementary charge)
   • Mass $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$

   The radius for a proton is: $r_p = \frac{m_p v}{eB}$

2. α-Particle:

   • Charge $q_{\alpha} = 2e$ (since it has two protons)
   • Mass $m_{\alpha} \approx 4 \times m_p$ (since it has four nucleons: 2 protons and 2 neutrons)

   The radius for an α-particle is: $r_{\alpha} = \frac{m_{\alpha} v}{q_{\alpha} B} = \frac{(4m_p) v}{2eB} = \frac{2 m_p v}{eB}$

Ratio of the Radii

Now we can find the ratio of the radii $\frac{r_p}{r_{\alpha}}$: $\frac{r_p}{r_{\alpha}} = \frac{\frac{m_p v}{eB}}{\frac{2 m_p v}{eB}} = \frac{m_p v}{eB} \cdot \frac{eB}{2 m_p v} = \frac{1}{2}$

Thus, the ratio of the radii of the circular paths described by the proton and the α-particle is: $\frac{r_p}{r_{\alpha}} = \frac{1}{2}$

This implies: $r_p : r_{\alpha} = 1 : 2$

Conclusion

Therefore, the correct answer is: A) 1 : 2

Why Other Options are Incorrect

• B) 1 : 4: This would suggest that the α-particle's radius is four times that of the proton, which contradicts our calculation that shows the α-particle's radius is actually larger but only twice that of the proton.
• C) 1 : 16: This option suggests an even larger discrepancy which is not supported by the physical reasoning or calculations.
• D) 4 : 1: This suggests that the proton has four times the radius of the α-particle, which is the opposite of what our calculations reveal.

Thus, only option A is consistent with the physics of the situation.