Question

A proton and an \( \alpha \)-particle enter with the same velocity \( \vec{v} \) in a uniform magnetic field \( \vec{B} \) such that \( \vec{v} \perp \vec{B} \). The ratio of the radii of their paths is:

• A. 2
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{4} \)
• D. 4

Hint:

Remember: the radius of circular motion in a magnetic field depends on the mass-to-charge ratio \( \left( \frac{m}{q} \right) \). A higher charge or lower mass leads to a smaller path radius.

Answer 1

The Correct Option is B

A charged particle moving perpendicularly into a uniform magnetic field undergoes a centripetal force from the Lorentz force, resulting in circular motion. The radius of this circular path is determined by the formula: \[r = \frac{mv}{qB}\] where: - \( m \) represents the particle's mass. - \( v \) is the particle's velocity. - \( q \) is the particle's charge. - \( B \) denotes the magnetic field strength. We will now calculate the radius for each particle: For the proton: - Mass \( m_p = m \) - Charge \( q_p = e \) \[r_p = \frac{mv}{eB}\] For the \( \alpha \)-particle (a helium nucleus): - Mass \( m_\alpha = 4m \) (composed of 2 protons and 2 neutrons) - Charge \( q_\alpha = 2e \) (due to 2 protons) \[r_\alpha = \frac{4m \cdot v}{2eB} = \frac{2mv}{eB}\] The ratio of the radii is calculated as follows: \[\frac{r_p}{r_\alpha} = \frac{\frac{mv}{eB}}{\frac{2mv}{eB}} = \frac{1}{2}\] Therefore, the ratio of the radii is: \[\text{Ratio of radii } = \frac{r_p}{r_\alpha} = \frac{1}{2}\]