Ionized hydrogen atoms and a-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths \( r_H: r_\alpha\) will be:

Question

A circular loop of radius 0.2 m carries a current of 4 A. What is the magnetic field at a point on the axis of the loop at a distance 0.2 m from the center?

Answer 1

The problem involves calculating the ratio of the radii of circular paths of two charged particles moving in a magnetic field, specifically ionized hydrogen atoms (protons) and alpha particles (He²⁺). Both particles enter the magnetic field perpendicularly, and their momenta are given as equal.

Key Concepts:

When a charged particle enters a magnetic field perpendicularly, it experiences a centripetal force that causes it to move in a circular path.
The formula relating the radius of the path \( r \) to the mass \( m \), velocity \( v \), charge \( q \), and magnetic field \( B \) is given by:

r = \frac{mv}{qB}

Given that the momenta (\( p = mv \)) for both particles are the same, we can express the radius as:

r = \frac{p}{qB}

Thus, the ratio of the radii of the paths of the ionized hydrogen atom (\( r_H \)) and the alpha particle (\( r_\alpha \)) is:

\frac{r_H}{r_\alpha} = \frac{\frac{p}{q_H B}}{\frac{p}{q_\alpha B}} = \frac{q_\alpha}{q_H}

Plugging in the charges for hydrogen and alpha particles:

Charge of ionized hydrogen atom (proton), \( q_H = +e \)
Charge of alpha particle, \( q_\alpha = +2e \)

Substituting these values, we get:

\frac{r_H}{r_\alpha} = \frac{2e}{e} = 2:1

This shows that the radius of the path of the ionized hydrogen atom is twice that of the alpha particle when both have the same momentum. Therefore, the correct answer is 2:1.

Answer 2