Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

Verified step by step guidance

1. To find the speed of the ions as they emerge from the velocity selector, use the condition for the velocity selector where the electric force equals the magnetic force: \( qE = qvB \). Here, \( E \) is the electric field, \( B \) is the magnetic field, and \( v \) is the velocity. Solve for \( v \) using the equation \( v = \frac{E}{B} \).
2. Substitute the given values for the electric field \( E = 155 \text{ V/m} \) and the magnetic field \( B = 0.0315 \text{ T} \) into the equation \( v = \frac{E}{B} \) to find the velocity of the ions.
3. For part (b), use the formula for the radius of the circular path in a magnetic field: \( r = \frac{mv}{qB} \), where \( r \) is the radius, \( m \) is the mass of the ion, \( v \) is the velocity found in part (a), \( q \) is the charge of the ion, and \( B \) is the magnetic field in the second region.
4. Rearrange the formula to solve for the mass \( m \): \( m = \frac{rqB}{v} \).
5. Substitute the known values: \( r = 0.175 \text{ m} \), \( B = 0.0175 \text{ T} \), and the velocity \( v \) from part (a) into the equation \( m = \frac{rqB}{v} \) to find the mass of the ions.

Key Concepts

Velocity Selector

A velocity selector uses perpendicular electric and magnetic fields to filter particles based on their velocity. The electric field exerts a force on charged particles, while the magnetic field applies a perpendicular force. Only particles with a specific velocity, where the electric and magnetic forces balance each other, pass through without deflection. This velocity is given by v = E/B, where E is the electric field strength and B is the magnetic field strength.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. It is the sum of the electric force (qE) and the magnetic force (qvB), where q is the charge, v is the velocity, E is the electric field, and B is the magnetic field. In a velocity selector, the Lorentz force is used to determine the velocity of particles that pass through without deflection.

Circular Motion in Magnetic Fields

When charged particles enter a magnetic field perpendicular to their velocity, they experience a centripetal force that causes them to move in a circular path. The radius of this path is determined by the balance between the magnetic force and the centripetal force, given by r = mv/qB, where m is the mass, v is the velocity, q is the charge, and B is the magnetic field strength. This relationship allows for the determination of particle mass when other variables are known.