Problem 9 - University Physics with Modern Physics

Question

At the instant when the current in an inductor is increasing at a rate of \(0.0640 \mathrm{A} / \mathrm{s},\) the magnitude of the self-induced emf is 0.0160 \(\mathrm{V}\) (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is 0.720 \(\mathrm{A} ?\)

Step-by-Step Solution

Answer

(a) The inductance is 0.25 H. (b) The magnetic flux through each turn is approximately 4.5 × 10^{-4} Wb.

Step 1: Understanding the emf Formula

The formula for induced emf \( \varepsilon \) in a coil is given by \( \varepsilon = -L \frac{di}{dt} \), where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate of change of current. Here, \( \varepsilon = 0.0160 \) V and \( \frac{di}{dt} = 0.0640 \) A/s. Our task is to find \( L \).

Step 2: Solving for Inductance

Rearranging the formula \( \varepsilon = -L \frac{di}{dt} \) gives us \( L = -\frac{\varepsilon}{\frac{di}{dt}} \). Substituting the values, we have \( L = -\frac{0.0160}{0.0640} \). Calculating this gives \( L = 0.25 \) H (Henrys).

Step 3: Understanding Magnetic Flux in a Solenoid

The magnetic flux through a single loop is given by \( \Phi = \frac{BA}{N} \), but for a solenoid, the relevant expression involves the current and inductance, \( L = N \frac{\Phi}{I} \). We have \( N = 400 \) turns, and the current \( I = 0.720 \) A, with the previously found \( L = 0.25 \) H.

Step 4: Solving for Magnetic Flux

Rearrange the formula \( L = N \frac{\Phi}{I} \) to find \( \Phi = \frac{L \cdot I}{N} \). Substituting the known values, \( \Phi = \frac{0.25 \times 0.720}{400} \), we find \( \Phi \approx 4.5 \times 10^{-4} \, \text{Wb} \) (Weber) per turn.

Key Concepts

Inductance

Inductance is a fundamental concept in electromagnetism, describing a coil's ability to resist changes in electric current. It acts similarly to inertia in mechanical systems. The higher the inductance, the more a coil opposes changes in current. It is measured in Henry (H). When you have an inductor, like a solenoid, and the current changes, the inductor will generate an electromotive force (emf) opposing that change.

This phenomenon leads to the creation of a magnetic field around the coil. The inductance, symbolized by \( L \), is determined by factors such as the number of turns in the coil, the coil's cross-sectional area, and the material inside the coil. For example, in our problem, using the formula \( \varepsilon = -L \frac{di}{dt} \), we found the inductance as \( L = 0.25 \) H by using the given change rate in current and the induced emf.

Inductance is crucial in designing circuits, particularly those used in transformers and inductors.

Self-induced emf

Self-induced emf is a property of a coil where a change in the electric current flowing through it induces an emf in the same coil. This self-induction is described by Lenz's law, which states that the induced emf always works against the change in current that created it.

• The self-induced emf, \( \varepsilon \), can be calculated with the formula \( \varepsilon = -L \frac{di}{dt} \), where \( L \) is the inductance and \( \frac{di}{dt} \) is the rate at which the current changes.
• The negative sign indicates that the induced emf opposes the change in current, aligning with Lenz's Law.

Understanding self-induced emf helps in analyzing how inductors behave in circuits and calculating the energy stored within a magnetic field. As in our exercise, the self-induced emf aligns directly with how quickly the current is changing, and the inductor's ability to resist that change.

Magnetic Flux

Magnetic flux represents the total magnetic field passing through a specific area, conceptualized as magnetic lines making their way through a loop or coil. It is measured in Weber (Wb). High magnetic flux means a strong magnetic field presence through the coil's surface.

For solenoids, magnetic flux is interconnected with current and number of turns via the inductor's formula. For instance, the inductance \( L = 0.25 \) H, number of turns \( N = 400 \), and current \( I = 0.720 \) A guide us in finding the flux through each loop:

• The formula \( L = N \frac{\Phi}{I} \) helps in determining magnetic flux.
• Rearranging to \( \Phi = \frac{L \cdot I}{N} \), gives us an average flux of approximately \( 4.5 \times 10^{-4} \) Wb per turn.

Thus, magnetic flux connects electric currents, magnetic fields, and coil geometry, integral to device functionality from transformers to sensors.