Question

Assume that an electron is describing a circular orbit of radius \(a\) about a positively charged nucleus. (a) By selecting an appropriate current and area, show that the equivalent orbital dipole moment is \(e a^{2} \omega / 2\), where \(\omega\) is the electron's angular velocity. \((b)\) Show that the torque produced by a magnetic field parallel to the plane of the orbit is \(e a^{2} \omega B / 2 .(c)\) By equating the Coulomb and centrifugal forces, show that \(\omega\) is \(\left(4 \pi \epsilon_{0} m_{e} a^{3} / e^{2}\right)^{-1 / 2}\), where \(m_{e}\) is the electron mass. \((d)\) Find values for the angular velocity, torque, and the orbital magnetic moment for a hydrogen atom, where \(a\) is about \(6 \times 10^{-11} \mathrm{~m} ;\) let \(B=0.5 \mathrm{~T}\).

Short answer

Question: Calculate the angular velocity, torque, and orbital magnetic moment for an electron orbiting a positively charged nucleus in a circular orbit of radius 6 x 10^-11 m, subjected to a magnetic field of 0.5 T parallel to the plane of the orbit. Answer: The angular velocity of the electron is approximately 3.24 x 10^15 s^-1, the torque on the orbiting electron is about 9.29 x 10^-25 Nm, and the orbital magnetic moment is approximately 1.86 x 10^-24 Am^2.

Step-by-step solution

(a) Determine the equivalent orbital dipole moment

We know that the magnetic dipole moment of a current loop is given by \(\boldsymbol{μ} = I \boldsymbol{A}\) where \(I\) is the current, and \(\boldsymbol{A}\) is the area vector (\(\boldsymbol{A} = A\boldsymbol{n}\)). In this case, the electron describes a circular orbit of radius \(a\) with angular velocity \(\omega\), so its current is \(I=\frac{e}{T}\), where \(T\) is the orbital period and \(e\) is the charge of the electron. We have, \(T=\frac{2\pi}{\omega}\), thus, \(I = \frac{e \omega}{2\pi}\). Next, the area of the circular orbit is \(A=\pi a^{2}\), and thus the area vector \(\boldsymbol{A} = \pi a^{2}\boldsymbol{n}\). Substituting these results, we obtain the equivalent orbital dipole moment: \(\boldsymbol{μ} = I \boldsymbol{A} = \left(\frac{e \omega}{2 \pi}\right)(\pi a^{2} \boldsymbol{n}) = \frac{ea^{2}\omega}{2}\boldsymbol{n}\).

(b) Calculate the torque on the orbiting electron

To find the torque produced by the magnetic field, we can use the formula: \(\boldsymbol{τ} = \boldsymbol{μ} × \boldsymbol{B}\), where \(\boldsymbol{B}\) is the magnetic field vector parallel to the plane of the orbit. Given that the magnetic field is parallel to the orbit plane, both \(\boldsymbol{μ}\) and \(\boldsymbol{B}\) are orthogonal to \(\boldsymbol{n}\), so their cross product simplifies as: \(\boldsymbol{τ} = \left| \boldsymbol{μ} \right| \left| \boldsymbol{B} \right| \boldsymbol{n} = \frac{ea^{2}\omega}{2}B\boldsymbol{n}\).

(c) Derive expression for angular velocity

To find the angular velocity (ω) of the orbiting electron, we need to equate the Coulomb and centrifugal forces acting on the electron: \(F_{coulomb} = F_{centrifugal}\) The Coulomb force acting on the electron is given by: \(F_{coulomb} = \frac{e^{2}}{4 \pi \epsilon_{0} a^{2}}\) The centrifugal force acting on the electron is given by: \(F_{centrifugal} = m_{e} a \omega^{2}\) Thus, equating both forces, we have: \(\frac{e^{2}}{4 \pi \epsilon_{0} a^{2}} = m_{e} a \omega^{2}\) Solving for \(\omega\), we obtain: \(\omega = \sqrt{\frac{e^2}{4\pi\epsilon_0 m_e a^3}} = \left(\frac{4 \pi \epsilon_{0} m_{e} a^{3}}{e^{2}}\right)^{-1 / 2}\).

(d) Calculate values for hydrogen atom

For a hydrogen atom, we are given that \(a = 6 \times 10^{-11} \ \text{m}\) and \(B = 0.5 \ \text{T}\). (a) Angular velocity: Using the expression for \(\omega\) obtained in (c), we have: \(\omega = \left(\frac{4 \pi \epsilon_{0} m_{e} a^{3}}{e^{2}}\right)^{-1 / 2} = \left(\frac{4 \pi \epsilon_{0} m_{e} (6 \times 10^{-11})^{3}}{(1.6 \times 10^{-19})^{2}}\right)^{-1 / 2} \approx 3.24 \times 10^{15} \ \text{s}^{-1}\). (b) Torque: Using the expression for \(\boldsymbol{τ}\) obtained in (b), we have: \(\boldsymbol{τ} = \frac{ea^{2}\omega}{2}B\boldsymbol{n} = \frac{(1.6 \times 10^{-19})(6 \times 10^{-11})^{2}(3.24 \times 10^{15})}{2}(0.5)\boldsymbol{n} \approx 9.29 \times 10^{-25} \ \mathrm{Nm} \boldsymbol{n}\). (c) Orbital magnetic moment: From the expression in (a), we already obtained the equivalent orbital dipole moment: \(\boldsymbol{μ} = \frac{ea^{2}\omega}{2}\boldsymbol{n} = \frac{(1.6 \times 10^{-19})(6 \times 10^{-11})^{2}(3.24 \times 10^{15})}{2}\boldsymbol{n} \approx 1.86 \times 10^{-24} \ \text{Am}^{2} \boldsymbol{n}\). We have now calculated the values for angular velocity, torque, and orbital magnetic moment for a hydrogen atom given the specific orbit radius and magnetic field strength.

Key concepts

Dipole Moment

The concept of dipole moment is central to understanding electromagnetic interactions. In simple terms, it represents the strength of a magnetic field generated by a loop of electric current. When dealing with atomic models, such as electrons in orbit, this concept helps in relating the movement of charges to magnetic effects. For a circular orbit of radius \( a \), an electron moving with angular velocity \( \omega \) generates a dipole moment.

• Current \( I \) can be calculated by considering the periodic motion of the electron, which equals the charge \( e \) divided by the orbital period \( T \).
• The area \( A \) through which this current travels is \( \pi a^2 \), which is the area of the circle made by the electron's path.

Combining these, the equation \( \mu = \frac{ea^2\omega}{2} \) is found. This defining relationship connects the motion of the electron to its magnetic properties.Understanding this concept aids in determining how atomic orbits contribute to material's magnetic properties, such as ferromagnetism or diamagnetism.

Magnetic Torque

Torque in magnetism explains how magnetic fields influence magnetic moments. When an electron orbits in a magnetic field \( B \), the field tries to align the magnetic moment \( \mu \) created by the orbiting electron. This alignment creates a force, called torque, that's orthogonal to both the magnetic moment and the magnetic field.The mathematical representation includes:

• Torque \( \tau = \mu \times B \), where the cross-product signifies the perpendicular nature of these vectors.
• This simplifies to \( \tau = \frac{ea^2\omega}{2}B \) when the magnetic field is in the plane of the orbit.

This simplification is because of the geometrical alignment where the vectors are primarily affecting each other at right angles, maximizing the torque. This understanding is crucial in designing devices like torque meters and engines utilizing magnetic fields.

Angular Velocity

Angular velocity \( \omega \) describes how fast an object rotates around a point or axis. In the context of electrons in orbit, it's about how quickly an electron sweeps around its nucleus over time. Understanding angular velocity is important for predicting the behavior of rotating bodies in fields.Calculation of angular velocity involves equating forces:

• The Coulomb force acting between the nucleus and electron \( F_{coulomb} = \frac{e^{2}}{4 \pi \epsilon_{0} a^{2}} \).
• Centrifugal force counteracting this as the electron moves in its orbit \( F_{centrifugal} = m_{e} a \omega^{2} \).

Equating these forces gives the relation \( \omega = \left(\frac{4 \pi \epsilon_{0} m_{e} a^{3}}{e^{2}}\right)^{-1/2} \) for an electron's angular velocity in its orbit. Understanding \( \omega \) is vital for calculations involving rotational dynamics and electron interactions.

Coulomb Force

The Coulomb force is the electric force of attraction or repulsion between charged particles. For an electron orbiting a positively charged nucleus, this force pulls the electron toward the nucleus.It is described by Coulomb's law as:

• \( F_{coulomb} = \frac{k \, q_1 \, q_2}{r^2} \), where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them.
• In this context, it simplifies to \( F_{coulomb} = \frac{e^{2}}{4 \pi \epsilon_{0} a^{2}} \) for an electron-nucleus interaction.

Coulomb's force governs atomic structures, dictating how electrons maintain their orbiting paths due to this constant pull. This force is fundamental to understanding how atoms remain stable yet reactive in nature.

Centripetal Force

Centripetal force is necessary to keep an object moving in a circular path. It's always directed towards the center of the circle, and in atomic systems, it balances the outward force due to rotational motion.This fundamental force is expressed as:

• \( F_{centrifugal} = m_{e} a \omega^2 \), where \( m_{e} \) is the mass of the electron and \( \omega \) its angular velocity.

Centripetal force counterbalances the Coulomb force, ensuring the electron remains in its circular orbit. This equilibrium condition is crucial for sustaining orbits without the electron spiraling into or escaping from the nucleus. Such a balance permits stable atomic models, lending insight into molecular formations and reactions.