Problem 4 - University Physics with Modern Physics

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 \(\mathrm{cm}\) long and has a diameter of 2.00 \(\mathrm{cm} .\) At a certain time, the current in the inner solenoid is 0.120 \(\mathrm{A}\) and is increasing at a rate of \(1.75 \times 10^{3} \mathrm{A} / \mathrm{s}\) . 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.

Step-by-Step Solution

Answer

(a) 4.75 x 10^-9 Wb; (b) 1.19 x 10^-5 H; (c) -0.0208 V.

Step 1: Calculate the Magnetic Field of the Inner Solenoid

The magnetic field inside a solenoid is given by the formula: \[ B = \mu_0 \times n \times I \]where \( \mu_0 \) is the permeability of free space \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T\cdot m/A} \), \( n \) is the number of turns per unit length \( n = \frac{25}{0.25 \ \mathrm{m}} \), and \( I \) is the current \( 0.120 \ \mathrm{A} \). Calculating \( n \): \[ n = \frac{25}{0.25} = 100 \ \mathrm{turns/m} \]Calculate \( B \):\[ B = 4\pi \times 10^{-7} \times 100 \times 0.120 = 1.51 \times 10^{-5} \ \mathrm{T} \]

Step 2: Find the Average Magnetic Flux Through Each Turn

The magnetic flux \( \Phi \) through one turn is given by the formula: \[ \Phi = B \times A \]where \( A \) is the cross-sectional area of the solenoid \( A = \pi \times \left(\frac{0.02}{2}\right)^2 \).Calculate \( A \):\[ A = \pi \times (0.01)^2 = \pi \times 10^{-4} \ = 3.14 \times 10^{-4} \ \mathrm{m^2} \]Calculate \( \Phi \):\[ \Phi = 1.51 \times 10^{-5} \times 3.14 \times 10^{-4} = 4.75 \times 10^{-9} \ \mathrm{Wb} \]

Step 3: Calculate the Mutual Inductance

The mutual inductance \( M \) is given by:\[ M = \frac{\Phi_{outer} \times N_{outer}}{I_{inner}} \]where \( N_{outer} = 300 \) and \( \Phi_{outer} = 4.75 \times 10^{-9} \ \mathrm{Wb} \).Since it's the same flux through each turn, the total flux for the outer coil:\[ \Phi_{total} = \Phi \times N_{outer} \]\[ \Phi_{total} = 4.75 \times 10^{-9} \times 300 = 1.425 \times 10^{-6} \ \mathrm{Wb} \]Thus,\[ M = \frac{1.425 \times 10^{-6}}{0.120} = 1.19 \times 10^{-5} \ \mathrm{H} \]

Step 4: Calculate the Induced EMF in the Outer Solenoid

The induced EMF \( \mathcal{E} \) in the outer solenoid is given by: \[ \mathcal{E} = -M \frac{dI}{dt} \]where \( \frac{dI}{dt} = 1.75 \times 10^3 \ \mathrm{A/s} \).Calculate \( \mathcal{E} \):\[ \mathcal{E} = -1.19 \times 10^{-5} \times 1.75 \times 10^3 = -0.0208 \ \mathrm{V} \]

Key Concepts

Solenoid

A solenoid is a long coil of wire, typically wound in many turns, that produces a magnetic field when an electric current passes through it. Think of it like a stack of loops arranged like a spring or a slinky.

• **Structure**: It consists of multiple loops or turns of wire wrapped tightly in a cylindrical shape.
• **Length & Diameter**: The solenoid in the original problem is 25 cm long and has a diameter of 2 cm.
• **Turns**: This solenoid specifically has 25 turns.

The main goal of using a solenoid is to create a uniform magnetic field inside it when electricity flows through. The field is strongest at the center and diminishes towards the edges.

Magnetic Flux

Magnetic flux represents the quantity of magnetic field passing through a given area. It provides a measure of how much of a magnetic field 'flows through' a surface.

• **Formula**: The formula for magnetic flux is given by \( \Phi = B \times A \), where \( B \) is the magnetic field strength and \( A \) is the area through which the field lines pass.
• **Area Calculation**: In the problem, the area of the solenoid's cross-section is calculated using the formula for the area of a circle: \( A = \pi \times \left(\frac{0.02}{2}\right)^2 \), resulting in an area of \( 3.14 \times 10^{-4} \ \mathrm{m^2} \).
• **Flux Calculation**: By substituting the calculated area and the magnetic field strength, the flux through each turn of the solenoid is \( 4.75 \times 10^{-9} \ \mathrm{Wb} \) (Weber).

The concept of magnetic flux is important because it plays a vital role in determining how electromagnetic interactions take place in this scenario.

Induced EMF

Induced electromotive force (EMF) refers to the voltage generated by a change in magnetic flux within a coil. It represents the influence a changing magnetic field has on an electric circuit, creating a current flow.

• **Faraday's Law**: This principle is based on Faraday's Law of electromagnetic induction, which states that an electromotive force is induced when there is a change in magnetic flux through a circuit.
• **Formula**: In the exercise, the induced EMF in the outer solenoid is calculated using \( \mathcal{E} = -M \frac{dI}{dt} \), where \( M \) is the mutual inductance and \( \frac{dI}{dt} \) is the rate of change of current.
• **Application**: For the given problem, the induced EMF becomes \( -0.0208 \ \mathrm{V} \), revealing how flux changes from the inner coil are communicated to the outer coil.

Induced EMF is essential in understanding how coils and solenoids can transfer magnetic energy, crucial for transformers, generators, and many other electrical devices.