Question

A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns (see Example 30.1). The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is 0.120 A and is increasing at a rate of \(1.75 \times 10^3\) A/s. For this time, calculate: (a) the average magnetic flux through each turn of the inner solenoid; (b) the mutual inductance of the two solenoids; (c) the emf induced in the outer solenoid by the changing current in the inner solenoid.

Short answer

(a) \(4.74 \times 10^{-9} \ \text{Wb}\), (b) \(9.42 \times 10^{-6} \ \text{H}\), (c) \(-0.0165\ \text{V}\).

Step-by-step solution

Calculate the Magnetic Field Inside the Inner Solenoid

First, we use the formula for the magnetic field inside a solenoid: \[ B = \mu_0 \cdot n \cdot I \]where \( \mu_0 = 4\pi \times 10^{-7} \ \text{T m/A} \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current. The number of turns per unit length is \( n = \frac{N}{L} = \frac{25}{0.25} \ \text{turns/m} \).

Compute the Magnetic Field

Plug in the values for \( n \) and \( I \):\[ n = \frac{25}{0.25} = 100 \ \text{turns/m} \]Using \( I = 0.120 \ \text{A} \), the magnetic field \( B \) becomes:\[ B = 4\pi \times 10^{-7} \cdot 100 \cdot 0.120 = 1.51 \times 10^{-5} \ \text{T} \].

Calculate the Average Magnetic Flux through Each Turn

The average magnetic flux \( \Phi \) through each turn is given by:\[ \Phi = B \cdot A \]where \( A \) is the cross-sectional area of the solenoid. The area \( A = \pi (0.01)^2 \ \text{m}^2 \) (since the diameter is 2.00 cm). Thus:\[ A = \pi \times (0.01)^2 = 3.14 \times 10^{-4} \ \text{m}^2 \].Now, \[ \Phi = 1.51 \times 10^{-5} \cdot 3.14 \times 10^{-4} = 4.74 \times 10^{-9} \ \text{Wb} \].

Determine the Mutual Inductance

The mutual inductance \( M \) between two coils is given by the formula:\[ M = \mu_0 \cdot n_1 \cdot n_2 \cdot \pi r^2 \cdot l \],where \( n_1 = 100 \ \text{turns/m} \), \( n_2 = \frac{300}{0.25} = 1200 \ \text{turns/m} \), \( r = 0.01 \ \text{m} \), and \( l = 0.25 \ \text{m} \). Inserting these values:\[ M = 4\pi \times 10^{-7} \cdot 100 \cdot 1200 \cdot \pi \times (0.01)^2 \cdot 0.25 \] results in:\[ M = 9.42 \times 10^{-6} \ \text{H} \].

Find the Induced EMF in the Outer Solenoid

The electromotive force (emf) induced in a solenoid by a changing current is given by:\[ \text{emf} = -M \frac{dI}{dt} \].Given \( \frac{dI}{dt} = 1.75 \times 10^3 \ \text{A/s} \), the emf is:\[ \text{emf} = -9.42 \times 10^{-6} \cdot 1.75 \times 10^3 \].Therefore, \[ \text{emf} = -0.0165 \ \text{V} \].

Key concepts

Magnetic Flux

Magnetic flux, often denoted by the symbol \( \Phi \), represents the total magnetic field passing through a given area. Think of it as the number of magnetic field lines crossing a specific surface. For a solenoid, this can be calculated using:\[ \Phi = B \cdot A \]Here, \( B \) is the magnetic field, and \( A \) is the cross-sectional area of the solenoid. The magnetic flux is crucial because it's the factor that changes when the current in the solenoid changes, leading to electromagnetic induction according to Faraday's Law. In this exercise, the average magnetic flux through each turn of the inner solenoid was derived using the solenoid's specific dimensions and current values.

Mutual Inductance

Mutual inductance is a measure of how the change in current in one coil induces an electromotive force (emf) in a nearby coil. It's a central concept in understanding how transformers operate. The mutual inductance \( M \) between two coils is given by the formula:\[ M = \mu_0 \cdot n_1 \cdot n_2 \cdot \pi r^2 \cdot l \]where:

• \( \mu_0 \) is the permeability of free space,
• \( n_1 \) and \( n_2 \) are the number of turns per unit length for each coil,
• \( r \) is the radius of the solenoid,
• \( l \) is the length of the solenoid.

The higher the mutual inductance, the more effective one coil is at inducing an emf in the other. This exercise calculated the mutual inductance of the two solenoids, demonstrating this principle.

Induced EMF

The induced electromotive force (emf) is the voltage generated by a change in magnetic flux over time, due to the principle of electromagnetic induction. According to Faraday's Law, the emf is calculated as:\[ \text{emf} = -M \frac{dI}{dt} \]where:

• \( M \) is the mutual inductance of the coils,
• \( \frac{dI}{dt} \) is the rate of change of current with time.

The negative sign in Faraday's Law indicates that the induced emf acts in a direction to oppose the change in current, a principle known as Lenz's Law. In this problem, we used the given rate of change of current to find the emf induced in the outer solenoid due to the inner solenoid.

Solenoid

A solenoid is a coil of wire that is often used to generate a magnetic field when an electric current passes through it. Solenoids are prevalent in various applications, such as electromagnets, inductors, and actuators. They consist of numerous turns of wire, which amplify the magnetic field produced by a given current. Key characteristics include:

• The number of turns of wire (\( N \)),
• The length of the solenoid (\( L \)),
• The coil's diameter or radius (\( r \)).

In our exercise, we dealt with two solenoids, with the inner coil having a smaller number of turns. The properties of the solenoid such as the number of turns per unit length and its dimensions were essential in calculating both the magnetic field and the mutual inductance. This highlights the role of solenoids in influencing the magnetic and induced electrical properties in circuits.