Question

An electron in a magnetic field moves counterclockwise on a circle in the \(x y\) -plane, with a cyclotron frequency of \(\omega=1.20 \cdot 10^{12} \mathrm{~Hz}\). What is the magnetic field, \(\vec{B}\) ?

Short answer

Answer: The magnitude of the magnetic field is approximately \(6.80 \cdot 10^{-4}\) Tesla.

Step-by-step solution

Write down the known values

The given values are: - Cyclotron frequency (\(\omega\)): \(1.20 \cdot 10^{12} \mathrm{Hz}\) - Electron charge (\(e\)): \(1.60 \cdot 10^{-19} \mathrm{C}\) - Electron mass (\(m_e\)): \(9.11 \cdot 10^{-31}\mathrm{kg}\)

Write down the formula for cyclotron frequency

The cyclotron frequency formula is: $$\omega = \frac{eB}{m_e}$$ where: \(\omega\) - cyclotron frequency \(e\) - electron charge \(B\) - magnitude of the magnetic field \(m_e\) - electron mass

Solve for the magnetic field (\(B\))

We want to find the magnitude of the magnetic field (\(B\)) given the cyclotron frequency (\(\omega\)), electron charge (\(e\)), and electron mass (\(m_e\)). Rearrange the formula to solve for \(B\): $$B = \frac{\omega m_e}{e}$$

Substitute known values and solve

Now, substitute the known values and solve for \(B\): $$B = \frac{(1.20 \cdot 10^{12}\mathrm{Hz})(9.11 \cdot 10^{-31}\mathrm{kg})}{1.60 \cdot 10^{-19} \mathrm{C}}$$ $$B \approx 6.80 \cdot 10^{-4} \mathrm{T}$$ So, the magnetic field \(\vec{B}\) has a magnitude of approximately \(6.80 \cdot 10^{-4}\) Tesla.

Key concepts

Magnetic Field

The concept of a magnetic field is central to understanding many physical phenomena, particularly how charged particles move. Imagine a magnetic field as a region in space where a magnetic force can be felt. This field is represented by field lines that emerge from the north pole of a magnet and enter at the south pole.

But it's not just bar magnets that create these fields. The Earth itself acts as a gigantic magnet, and even the movement of charges in a wire can generate a magnetic field. In our daily lives, magnetic fields are also crucial in devices like MRI machines, electric motors, and the Earth's protection against solar winds.

Magnetic fields are vector fields, meaning they have both magnitude and direction, symbolized by the vector \( \vec{B} \). The strength of the field is indicated by the field's magnitude, usually measured in Teslas (T) in the International System of Units (SI).

Electron Motion in Magnetic Field

Now let's zoom into the motion of an electron in a magnetic field. When an electron, which carries a charge, enters a magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field vector. This force doesn't change the speed of the electron but alters its direction, causing it to move along a circular or helical path.

Cyclotron Motion

This circular motion is known as cyclotron motion, characterized by the cyclotron frequency \( \omega \). The cyclotron frequency is crucial for understanding the dynamics of charged particles in a magnetic field, and it's given by the formula \( \omega = \frac{eB}{m_e} \), where \(e\) is the charge of the electron, \(B\) is the magnetic field strength, and \(m_e\) is the electron's mass. The resulting circular path has a radius called the 'gyroradius' and is determined by the velocity of the particle and the strength of the magnetic field. Because of these principles, we can manipulate electron paths using magnetic fields, which is the underlying principle of devices like cyclotrons, used to accelerate charged particles.

Calculating Magnetic Field Strength

Calculating the magnetic field strength is critical for understanding the environment in which charged particles move. To find the magnetic field from the frequency of electron motion, we use the rearranged cyclotron frequency formula \( B = \frac{\omega m_e}{e} \).

For our exercise, all necessary quantities were identified, including the cyclotron frequency \( \omega \), the electron charge \( e \), and the electron mass \( m_e \). By substituting these values into the formula, we calculated the magnetic field strength. In practical applications, engineers and scientists use similar calculations to design and operate equipment such as magnetic resonance imaging (MRI) machines and particle accelerators. Understanding how to calculate this field strength is not just a textbook exercise – it's a fundamental skill that bridges theory with real-world technology.