Question

An electron with a speed of \(4.00 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) enters a uniform magnetic field of magnitude \(0.0400 \mathrm{~T}\) at an angle of \(35.0^{\circ}\) to the magnetic field lines. The electron will follow a helical path.

Short answer

Answer: The radius of the helix followed by the electron is \(3.42 × 10^{-4} \, \mathrm{m}\), and the pitch of the helix is \(2.06 × 10^{-4} \, \mathrm{m}\).

Step-by-step solution

Determine the given values

The given values in the exercise are as follows: - Speed of the electron, \(v = 4.00 × 10^{5} \, \mathrm{m/s}\) - Magnetic field strength, \(B = 0.0400 \, \mathrm{T}\) - Angle between electron's velocity and magnetic field lines, \(\theta = 35.0^{\circ}\)

Calculate the velocity components parallel and perpendicular to the magnetic field

Since the angle between the electron's velocity and the magnetic field lines is given, we can calculate the components of the electron’s velocity parallel and perpendicular to the magnetic field. $$ v_{\parallel} = v \cos(\theta) $$ $$ v_{\perp} = v \sin(\theta) $$ First, convert the given angle to radians: $$ \theta = 35.0^{\circ} \cdot \frac{\pi}{180} = 0.610 \, \mathrm{rad} $$ Now, calculate the components of velocity: $$ v_{\parallel} = 4.00 × 10^{5} \, \mathrm{m/s} \cdot \cos(0.610) = 3.28 × 10^{5} \, \mathrm{m/s} $$ $$ v_{\perp} = 4.00 × 10^{5} \, \mathrm{m/s} \cdot \sin(0.610) = 2.40 × 10^{5} \, \mathrm{m/s} $$

Calculate the centripetal force and radius of the helix

The force on a moving charged particle in a magnetic field is given by: $$ F = q \cdot v_{\perp} \cdot B $$ Where \(q\) is the charge of the electron, which is equal to \(-1.60 × 10^{-19} \, \mathrm{C}\) (the negative sign indicates its charge is opposite to a proton). The centripetal force acting on the electron is due to the magnetic force: $$ F = \frac{m \cdot v_{\perp}^{2}}{r} $$ Where \(m\) is the mass of the electron, which is equal to \(9.11 × 10^{-31} \, \mathrm{kg}\), and \(r\) is the radius of the helix. Equating the two expressions for force, we get: $$ q \cdot v_{\perp} \cdot B = \frac{m \cdot v_{\perp}^{2}}{r} $$ Solving for \(r\), we find: $$ r = \frac{m \cdot v_{\perp}}{q \cdot B} = \frac{9.11 × 10^{-31} \, \mathrm{kg} \cdot 2.40 × 10^{5} \, \mathrm{m/s}}{1.60 × 10^{-19} \, \mathrm{C} \cdot 0.0400 \, \mathrm{T}} = 3.42 × 10^{-4} \, \mathrm{m} $$

Calculate the pitch of the helix

The pitch of the helix is the distance the electron travels parallel to the magnetic field during one complete turn of the helix. To calculate the pitch, we need the time it takes for the electron to complete one full turn, which is the period of the motion: $$ T = \frac{2 \pi m}{q B} $$ Now, we can calculate the pitch using the parallel velocity component and the period: $$ P = v_{\parallel} T = v_{\parallel} \frac{2 \pi m}{q B} = \frac{3.28 × 10^{5} \mathrm{m/s} \cdot 2 \pi (9.11 × 10^{-31} \, \mathrm{kg})}{1.60 × 10^{-19} \, \mathrm{C} \cdot 0.0400 \, \mathrm{T}} = 2.06 × 10^{-4} \, \mathrm{m} $$ Therefore, the radius of the helix followed by the electron is \(3.42 × 10^{-4} \, \mathrm{m}\), and the pitch of the helix is \(2.06 × 10^{-4} \, \mathrm{m}\).

Key concepts

Magnetic Force on Moving Charges

When a charged particle such as an electron moves through a magnetic field, it experiences a force known as the magnetic Lorentz force. This force is described by the equation:
\[ F = q(v \times B) \]
Here, \( F \) is the magnetic force, \( q \) is the electric charge of the particle, \( v \) is the particle’s velocity, and \( B \) is the magnetic field. The force is at its maximum when the particle's velocity is perpendicular to the magnetic field, and it is zero when the velocity is parallel to the field.

The direction of this force is perpendicular to both the velocity of the particle and the magnetic field lines, and it is given by the right-hand rule: if you point the thumb of your right hand in the direction of the particle's velocity and your fingers in the direction of the magnetic field, your palm will point in the direction of the force acting on a positively charged particle. An electron, being negatively charged, will feel the force in the opposite direction.

Therefore, it's crucial to distinguish between the parallel (\(v_{\parallel}\)) and perpendicular (\(v_{\perp}\)) velocity components in relation to the magnetic field to understand the nature of the electron's motion through the field.

Helical Path of Electrons

Electrons moving through a magnetic field often follow a spiral path known as a helix. This occurs because the magnetic force acts as a centripetal force, pulling the electron in a circular motion perpendicular to the magnetic field while it also continues to move forward. The combination of these motions—forward and circular—results in a helical trajectory.

The radius of this helix is determined by the balance between the magnetic force and the centripetal force required to maintain the circular portion of the motion. We can calculate the radius using the perpendicular component of the velocity and the magnetic field strength, as shown in the original exercise solution.

It's also interesting to note that the pitch of the helix— the linear advancement along the magnetic field per turn—depends on the time it takes to complete one revolution, called the period, and the component of the velocity that is parallel to the field. The pitch represents how far the electron will move along the magnetic field line while completing one full circular motion.

Centripetal Force

The concept of centripetal force is central to understanding the circular portion of the helical motion of electrons in a magnetic field. By definition, centripetal force is the inward force required for an object to follow a curved path. In this context, the specific force is magnetic, and it is given by the equation:
\[ F_{centripetal} = \frac{m v_{\perp}^2}{r} \]
Where \( m \) is the mass of the electron, \( v_{\perp} \) is the component of the electron's velocity that is perpendicular to the magnetic field, and \( r \) is the radius of the curved path (in this case, the radius of the helix).

Equating the magnetic force to the centripetal force provides us with a way to determine the radius of the helical path when the electron is moving through the magnetic field. The subtle balance between these forces is what leads to the stable helical motion of the electron.