Problem 53 - Fundamentals of Physics

Question

A long solenoid has 100 turns/cm and carries current \(i\). An electron moves within the solenoid in a circle of radius \(2.30 \mathrm{~cm}\) perpendicular to the solenoid axis. The speed of the electron is \(0.0460 c(c=\) speed of light \() .\) Find the current \(i\) in the solenoid.

Step-by-Step Solution

Answer

The current in the solenoid is approximately 1.15 A.

Step 1: Understanding the Problem

We are given a solenoid with 100 turns/cm and an electron moving in a circle perpendicular to the solenoid's axis, with a speed of 0.0460 times the speed of light. We need to find the current, \(i\), in the solenoid.

Step 2: Formulate Relevant Equations

For a solenoid, the magnetic field \(B\) inside is calculated as \(B = \mu_0 n i\) where \(n\) is the number of turns per unit length, and \(\mu_0\) is the permeability of free space \(\mu_0 = 4\pi \times 10^{-7} \ \text{T}\cdot\text{m/A}\). The centripetal force required to keep the electron in circular motion is \(F_c = \frac{m_ev^2}{r}\) and is provided by the magnetic force \(F_B = qvB\), where \(q\) is the charge of an electron and \(v\) is its velocity.

Step 3: Calculate Magnetic Field

Using \(B = \mu_0 n i\), with \(n = 10000 \ \text{turns/m} \) (since 100 turns/cm = 10000 turns/m). Thus, \(B = \mu_0 \times 10000 \times i\).

Step 4: Equate Forces for Circular Motion

The centripetal force is equal to the magnetic Lorentz force: \(\frac{m_ev^2}{r} = evB\). Substitute \(B\) from the magnetic field equation: \(\frac{m_ev^2}{r} = ev(\mu_0 n i)\).

Step 5: Solve for Current

Simplify and solve for \(i\): \[ i = \frac{m_ev}{e\mu_0 nr} \]Where: - \(m_e = 9.11 \times 10^{-31} \) kg (mass of electron),- \(v = 0.0460 \times c = 0.0460 \times 3.00 \times 10^8 \ \text{m/s}\),- \(e = 1.60 \times 10^{-19} \) C (charge of electron),- \(r = 2.30 \times 10^{-2} \ \text{m}\),- \(n = 10000 \ \text{turns/m}\).Calculate \(i\) after substituting these values.

Step 6: Perform Calculations

Substitute the given values into the equation:\[ i = \frac{9.11 \times 10^{-31} \times 0.0460 \times 3.00 \times 10^8}{1.60 \times 10^{-19} \times 4\pi \times 10^{-7} \times 10000 \times 2.30 \times 10^{-2}} \]Compute the result to find \(i\). After performing the calculations, \(i \approx 1.15 \ \text{A}\).

Key Concepts

Magnetic Field in Solenoid

A solenoid is essentially a coil of wire, and when an electric current passes through it, a magnetic field is created inside. This is the principle behind how solenoids work. The magnetic field inside a long solenoid can be calculated with the formula:

• \( B = \mu_0 n i \)

Here, \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space, \(n\) is the turn density (turns per unit length), and \(i\) is the current flowing through the solenoid.
In our problem, the solenoid has 100 turns per centimeter, which is 10000 turns per meter. This high density of turns amplifies the magnetic field created by the current, making solenoids efficient at generating powerful magnetic fields.
This magnetic field is uniform inside the solenoid and is key to determining how objects like electrons behave within it.

Centripetal Force and Magnetic Force

When an electron moves through a magnetic field perpendicularly, it experiences a force. This force is called the Lorentz force and is given by:

• Magnetic force: \( F_B = qvB \)

Where \(q\) is the charge of the electron, \(v\) is its velocity, and \(B\) is the magnetic field. Now, when the electron moves in a circular path, a centripetal force is needed to keep it on this path. This force is expressed as:

• Centripetal force: \( F_c = \frac{m_ev^2}{r} \)

The electron remains in its circular path because the magnetic force provides exactly the necessary centripetal force. In this problem, we equate these forces to analyze the motion:

• \( \frac{m_ev^2}{r} = qvB \)

By understanding this relationship, we can deduce other quantities, such as the current in the solenoid. Remember, the delicate balance between these forces is key for circular motion.

Electron Circular Motion

In a magnetic field, an electron can move in circles at a constant speed. This is a fascinating phenomenon, deeply reliant on the forces involved. The circular path happens because the magnetic force always acts perpendicular to the velocity of the electron.
As a result, the electron continuously changes direction while maintaining constant speed, tracing a circular trajectory. This kind of motion can be calculated and predicted using our earlier mentioned centripetal and magnetic forces. The radius of the circle, in this particular problem, is given as 2.30 cm.
Because of these properties, electrons moving in magnetic fields form the basis of many technologies like cyclotrons and mass spectrometers. These devices manipulate the paths of charged particles for various applications.

Equations of Motion for Charged Particle

The movement of charged particles, like electrons, in electric and magnetic fields, is governed by specific equations of motion. These equations consider the forces acting on the particle (in this case, primarily the Lorentz force).

• The general principle is expressed as \( F = ma \), where \(a\) is acceleration.

However, in a magnetic field, applied forces are perpendicular to the velocity of the particle, leading purely to changes in direction but not speed. This is why the electron moves in a circle with constant speed.
For a more specific motion, we use the force balance equation:

• \( \frac{m_ev^2}{r} = evB \)

This equation links the particle's mass \(m_e\), charge \(e\), velocity \(v\), magnetic field \(B\), and circle radius \(r\). Understanding this relationship helps solve for unknowns. In the problem at hand, it allows us to solve for the solenoid current \(i\).
By dealing with these equations, understanding the motion of charged particles becomes intuitive and enables the calculation of factors such as speed, force, and current in varied electromagnetic scenarios.