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

When a charged particle enters a uniform magnetic field perpendicular to its velocity, it experiences a centripetal force due to the Lorentz force, and follows a circular path. The radius of the circular motion is given by the formula: \[ r = \frac{mv}{qB} \] where: - \( m \) is the mass of the particle - \( v \) is the velocity - \( q \) is the charge - \( B \) is the magnetic field strength Let’s compute the radius for both particles: For the proton: - mass \( m_p = m \) - charge \( q_p = e \) \[ r_p = \frac{mv}{eB} \] For the \( \alpha \)-particle (which is a helium nucleus): - mass \( m_\alpha = 4m \) (2 protons + 2 neutrons) - charge \( q_\alpha = 2e \) (due to 2 protons) \[ r_\alpha = \frac{4m \cdot v}{2eB} = \frac{2mv}{eB} \] Now, take the ratio of the radii: \[ \frac{r_p}{r_\alpha} = \frac{\frac{mv}{eB}}{\frac{2mv}{eB}} = \frac{1}{2} \] Hence, \[ \text{Ratio of radii } = \frac{r_p}{r_\alpha} = \frac{1}{2} \]