Question

A singly ionized (one electron removed) \(^{40}\)K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform magnetic field (\(B'\)) oriented at right angles to their velocity. \(^{40}\)K contains 19 protons and 21 neutrons and has a mass of 6.64 \(\times\) 10\(^{-26}\) kg. (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of \(B'\) so that the ions will be bent into a semicircle of radius 12.5 cm?

Short answer

Electric field: 112.5 V/m; Magnetic field \(B'\): 0.0150 T.

Step-by-step solution

Understanding Velocity Selector

The velocity selector allows ions to pass through without deviation by balancing electric and magnetic forces. To achieve this, the electric field (\(E\)) and magnetic field (\(B\)) must equalize the ion's forces. The force due to the electric field is \(qE\) and due to the magnetic field is \(qvB\), where \(q\) is charge, \(v\) is velocity. For no deflection: \(qE = qvB\). Thus, \(E = vB\).

Calculate Electric Field

Given velocity \(v = 4.50 \text{ km/s} = 4500 \text{ m/s}\) and magnetic field \(B = 0.0250 \text{ T}\), we use \(E = vB\) to find the electric field: \[E = (4500 \text{ m/s}) \times (0.0250 \text{ T}) = 112.5 \text{ V/m}\].

Understanding Second Magnetic Field Effect

When the ion enters the second magnetic field \(B'\), it experiences a centripetal force causing circular motion. The magnetic force is \( qvB' \), which provides the centripetal force \( mv^2/r \). The radius of the circular path is given as 12.5 cm.

Calculate Second Magnetic Field

Solving for \(B'\) using \( qvB' = mv^2/r \), where \( r = 0.125 \text{ m}\), mass \( m = 6.64 \times 10^{-26} \text{ kg}\), and charge \( q = 1.6 \times 10^{-19} \text{ C}\) (since it is singly ionized), we have: \[ B' = \frac{mv}{qr} = \frac{(6.64 \times 10^{-26} \text{ kg}) \times (4500 \text{ m/s})}{(1.6 \times 10^{-19} \text{ C}) \times (0.125 \text{ m})} \approx 0.0150 \text{ T}\].

Conclusion on Field Magnitudes

Thus, the electric field strength required in the selector is 112.5 V/m, and the magnetic field \(B'\) for bending ions into a semicircle is approximately 0.0150 T.

Key concepts

Magnetic Fields

A magnetic field is a region where a magnetic force is exerted. It is created by moving electric charges and is described by the magnetic field vector, which is commonly denoted by \( B \). Magnetic fields have both magnitude and direction, and they can affect other moving charges.

In an ion velocity selector, charged particles move perpendicularly through both an electric and magnetic field. The magnetic component of the force on a moving charge (such as an ion) is given by the equation:

\( F = qvB \)

where \( F \) is the magnetic force, \( q \) is the charge of the particle, \( v \) is the velocity, and \( B \) is the magnetic field strength. The direction of the magnetic field relative to the movement of the charge is crucial, as the force depending on this will deflect the particle's motion. The ion's path will be altered unless other conditions or fields are adjusted to counterbalance these forces.

Electric Fields

An electric field exists around charged particles. It is represented by the electric field vector \( E \), and it indicates the direction and magnitude of the electric force a positive charge would experience. In a velocity selector, the relationship between electric and magnetic fields is crucial for allowing particles with a specific velocity to pass through without being deflected.

Within such a selector, the electric force, which is \( qE \), counters the magnetic force \( qvB \). This balance ensures that only particles moving at a velocity \( v \) determined by \( E = vB \) will pass through undeviated. This precise relationship allows the selection of particles based on their speed.

Centripetal Force

Centripetal force is necessary for an object to move in a circular path. It acts towards the center of the circle and keeps the object moving in that curved path rather than continuing in a straight line. For ions entering a magnetic field at right angles, this force is provided by the magnetic force.

When an ion enters a second magnetic field, the magnetic force acts as the centripetal force required to bend the ion into a circular path. The balance of forces is described by:

\( qvB' = \frac{mv^2}{r} \)

where \( m \) is the mass of the ion, \( v \) is its speed, \( B' \) is the magnetic field strength, and \( r \) is the radius of the path. Solving for \( B' \) helps determine the magnetic field strength needed to achieve a specific circular path radius.

Singly Ionized Atoms

Atoms become ionized when they gain or lose electrons. A singly ionized atom has either lost or gained one electron, resulting in a net charge. In the context of a velocity selector, understanding this charge helps calculate forces acting on the ion.

For a singly ionized ion, such as the \(^{40}\)K atom presented in the problem, its charge \( q \) equals one elementary charge, or \( 1.6 \times 10^{-19} \) C. The atomic mass and charge are pivotal in determining how the ion behaves under electric and magnetic fields, influencing calculations of field strengths, forces, and, consequently, the ion's path as it moves.