Problem 7 - 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) Inductance is 0.25 H. (b) Average magnetic flux is 0.00045 Wb.

Step 1: Identify Known Values

The rate of change of current \( \frac{di}{dt} = 0.0640 \, \text{A/s} \) and the self-induced emf \( \varepsilon = 0.0160 \, \text{V} \) are given. We need to find the inductance \( L \). We also have a solenoid with 400 turns, and the current is \( i = 0.720 \, \text{A} \).

Step 2: Use The Formula for Inductance

The magnitude of the self-induced emf is given by the formula \( \varepsilon = L \frac{di}{dt} \). We need to solve for \( L \).

Step 3: Solve for Inductance \( L \)

Rearrange the formula \( \varepsilon = L \frac{di}{dt} \) to \( L = \frac{\varepsilon}{\frac{di}{dt}} \). Substitute the known values: \( L = \frac{0.0160}{0.0640} \).

Step 4: Calculate Inductance \( L \)

Perform the division: \( L = \frac{0.0160}{0.0640} = 0.25 \, \text{H} \). Thus, the inductance of the inductor is \( 0.25 \, \text{H} \).

Step 5: Find Formula for Magnetic Flux

The magnetic flux \( \Phi \) through one turn is related to inductance by \( \Phi = \frac{L \cdot i}{N} \), where \( N \) is the number of turns and \( i \) is the current.

Step 6: Calculate Magnetic Flux

Substitute the known values into the formula: \( \Phi = \frac{0.25 \cdot 0.720}{400} \).

Step 7: Perform Calculation for \( \Phi \)

Calculate \( \frac{0.25 \cdot 0.720}{400} = 0.00045 \, \text{Wb} \). Thus, the average magnetic flux through each turn is \( 0.00045 \, \text{Wb} \).

Key Concepts

Self-induced emf

When we talk about self-induced emf, we're diving into the fascinating world of electromagnetism. This concept revolves around the idea that an electric current flowing through an inductor can induce an electromotive force (emf) within itself. This phenomenon occurs due to the changing magnetic field as the current varies.

• When the current flowing through an inductor changes, it creates a changing magnetic field.
• This change in the magnetic field induces a voltage, known as the self-induced emf.
• The direction of this induced emf always opposes the change in current, following Lenz's Law.

The relationship between self-induced emf (\( \varepsilon \)) and inductance (\( L \)) is given by the formula:
\( \varepsilon = L \frac{di}{dt} \)
where \( \frac{di}{dt} \) is the rate of change of current. This formula implies that higher inductance or faster changes in current result in a greater self-induced emf.

Solenoid

A solenoid is one of the most common inductor forms. It's essentially a coil of wire, and when an electric current passes through, it produces a magnetic field.

• They are used in creating magnetic fields for a wide range of applications, from electronic circuits to electromagnets.
• The magnetic field inside a solenoid is often uniform and parallel to the axis of the coil when the solenoid is long compared to its diameter.
• The field strength can be increased by adding more turns of wire or increasing the current.

In our exercise, a solenoid with 400 turns is used. The inductance of a solenoid depends on factors such as the number of turns, the cross-sectional area, and the length of the coil. By rearranging the windings tightly, a stronger magnetic field is achieved, enhancing its ability to store energy in the magnetic field.

Magnetic Flux

Magnetic flux refers to the number of magnetic field lines passing through a given area. It's a critical concept when discussing how inductors work.

• The flux, often denoted by \( \Phi \), is a measure of the strength and distribution of a magnetic field over a certain area.
• It's measured in Webers (Wb).
• In a solenoid, the average magnetic flux through each loop can be calculated by using the formula: \( \Phi = \frac{L \cdot i}{N} \)

Here, \( L \) is the inductance, \( i \) is the current, and \( N \) is the number of turns. By applying this formula, one can determine the average magnetic flux through each loop, showing how the energy is stored in the magnetic field within the solenoid. Understanding magnetic flux helps us grasp the dynamics of energy transformation and storage within inductive systems.