Problem 14 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A 420-turn solenoid, \(25 \mathrm{~cm}\) long, has a diameter of \(2.5 \mathrm{~cm}\). A 15-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to \(5.0 \mathrm{~A}\) in \(0.60 \mathrm{~s}\), what will be the induced emf in the short coil during this time?

Step-by-Step Solution

Answer

The induced emf in the short coil is approximately \(-2.0 \mathrm{~mV}\).

Step 1: Understand Faraday's Law of Induction

Faraday's Law states that the induced electromotive force (emf) in a coil is equal to the negative rate of change of magnetic flux through the coil: \( \varepsilon = -\frac{d\Phi}{dt} \), where \( \Phi \) is the magnetic flux. The negative sign represents Lenz's Law, indicating the direction of the induced emf opposes the change in magnetic flux.

Step 2: Calculate Magnetic Flux through the Coil

The magnetic flux \( \Phi \) through a solenoid is given by \( \Phi = B \cdot A \cdot N \), where \( B \) is the magnetic field, \( A \) is the cross-sectional area, and \( N \) is the number of turns of the coil. Inside a solenoid, the magnetic field \( B \) is given by \( B = \mu_0 \frac{nI}{l} \), where \( n \) is the number of turns per unit length, \( I \) is the current, and \( \mu_0 \) is the permeability of free space \( \approx 4\pi \times 10^{-7} \mathrm{~Tm/A} \).

Step 3: Compute the Magnetic Flux Change

We need to find the change in magnetic flux \( \Delta \Phi = \Phi_{final} - \Phi_{initial} \). Initially, the current is 0, so \( \Phi_{initial} = 0 \). For \( \Phi_{final} \), use \( B_{final} = \mu_0 \cdot \frac{420}{0.25} \cdot 5.0 \). The cross-sectional area \( A = \pi \left( \frac{0.025}{2} \right)^2 \), and the coil has 15 turns, so \( \Phi_{final} = B_{final} \cdot A \cdot 15 \).

Step 4: Calculate Induced Emf using Faraday's Law

Apply Faraday's Law: \( \varepsilon = -\frac{\Delta \Phi}{\Delta t} \), where \( \Delta t = 0.6 \mathrm{~s} \). Calculate \( \varepsilon \) using \( \varepsilon = -\frac{B_{final} \cdot A \cdot 15}{0.6} \). Substitute values from the previous step to find the numerical value of \( \varepsilon \).

Step 5: Final Calculation

Substitute \( B_{final} = 4\pi \times 10^{-7} \cdot 1680 \cdot 5.0 \), \( A = \pi \times 0.0125^2 \), and calculate \( \Phi_{final} \) using the number of turns of the coil. Compute the induced emf: \( \varepsilon = -\frac{\Phi_{final}}{0.6} \).

Key Concepts

Electromotive Force (emf)

Electromotive force, commonly known as emf, is not a force as its name suggests. Instead, it's a measurement of the energy that causes current to flow through the circuit. It represents the "voltage" generated by a source, such as a battery or, in our exercise, the induction within a coil.

When a change occurs in the magnetic environment of a coil, an emf is produced, according to Faraday's Law of Induction. This law tells us that emf is directly related to the rate at which magnetic flux changes in a coil. A typical scenario in which emf is induced is when there's a change in the current flowing through a nearby solenoid, as described in our original exercise.

Faraday’s Law is mathematically stated as:

• \( \varepsilon = - \frac{d\Phi}{dt} \)

Here, \( \varepsilon \) is the induced emf, and \( \frac{d\Phi}{dt} \) is the rate of change of magnetic flux, indicating that faster changes produce higher emf. The negative sign represents Lenz's Law, signifying that the induced emf always works to oppose the change in magnetic flux that produces it.

Magnetic Flux

Magnetic flux, denoted as \( \Phi \), is the measure of the quantity of magnetism, taking into account the strength and the extent of the magnetic field. It is a crucial factor in the generation of electromotive force through the process of electromagnetic induction.

Imagine magnetic flux as "lines" passing through a certain area. The denser these lines, the stronger the magnetic field, and consequently, the larger the magnetic flux. Magnetic flux is dependent on three major attributes:

• The magnetic field strength \( B \)
• The area \( A \) through which the field lines pass
• The orientation of the field relative to that area

Mathematically, magnetic flux is given by:

• \( \Phi = B \cdot A \cdot \cos(\theta) \)

In our context, with a solenoid, the angle \( \theta \) is typically zero, simplifying the expression to \( \Phi = B \cdot A \), as the field lines pass directly through the cross-section of the coil. Inside the solenoid, the magnetic field leads to different amounts of flux change, affecting how much emf will be induced.

Solenoid

A solenoid is essentially a coil of wire, tightly wound in the form of a helix, that can produce a substantial magnetic field. Solenoids are greatly valued for their ability to convert electrical energy into mechanical action, and they play a significant role in many electronics and automation systems.

In our original exercise, the solenoid is described with specific parameters that determine its capacities, such as:

• The number of turns: In our case, 420 turns, which contributes to how strong the induced field can become.
• Length: The solenoid in the exercise is 25 cm long, affecting the field uniformity inside the solenoid.
• Diameter: With a diameter of 2.5 cm, which determines the cross-sectional area through which the magnetic field lines travel. This is crucial for calculating the magnetic flux.

Inside a solenoid, assuming current flows, a nearly uniform magnetic field \( B \) is generated. This magnetic field strength can be determined using the formula:

• \( B = \mu_0 \frac{nI}{l} \)

Here, \( n \) is the number of turns per unit length, \( I \) is the current, and \( \mu_0 \) is the permeability of free space. This uniform magnetic field is what causes the changes in magnetic flux that result in the induction of emf in a nearby coil when the current changes, as discussed in the problem.