Problem 43 - University Physics with Modern Physics

Question

One solenoid is centered inside another. The outer one has a length of 50.0 \(\mathrm{cm}\) and contains 6750 coils, while the coaxial inner solenoid is 3.0 \(\mathrm{cm}\) long and 0.120 \(\mathrm{cm}\) in diameter and contains 15 coils. The current in the outer solenoid is changing at 49.2 A/s. (a) What is the mutual inductance of these solenoids? (b) Find the emf induced in the innner solenoid.

Step-by-Step Solution

Answer

(a) 0.00369 H, (b) -0.1815 V

Step 1: Understand the Concept of Mutual Inductance

Mutual inductance ( M ) between two coils relates the emf induced in one coil due to the change of current in the other. It depends on the number of turns, the geometry, and magnetic permeability.

Step 2: Calculate the Magnetic Field of Outer Solenoid

The magnetic field inside a solenoid is given by the formula: \( B = \mu_0 \cdot n \cdot I \), 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. For the outer solenoid:\[ n = \frac{6750}{0.5} = 13500 \, \text{turns/m} \]Hence, \[ B = \mu_0 \cdot 13500 \cdot I \].

Step 3: Calculate Flux Through Inner Solenoid

The magnetic flux \( \Phi \) through the inner solenoid due to the outer solenoid's field is: \[ \Phi = B \cdot A \cdot N_{\text{inner}} \], where\( A \) is the cross-sectional area of the inner solenoid and \( N_{\text{inner}} = 15 \) is its number of turns. With radius \( r = 0.0012 \text{ m} \), we have: \[ A = \pi r^2 \approx 4.52 \times 10^{-6} \text{ m}^2 \], so \[ \Phi = (\mu_0 \cdot 13500 \cdot I) \cdot 4.52 \times 10^{-6} \cdot 15 \].

Step 4: Derive the Formula for Mutual Inductance

The mutual inductance \( M \) is given by: \[ M = \frac{\Phi}{I} \]. Using the flux expression from Step 3: \[ M = \mu_0 \cdot 13500 \cdot 4.52 \times 10^{-6} \cdot 15 \approx 0.00369 \, \text{H} \].

Step 5: Find Emf Induced in Inner Solenoid

The induced emf \( \varepsilon \) due to changing current is given by:\( \varepsilon = -M \frac{dI}{dt} \). Using the mutual inductance from Step 4 and \( \frac{dI}{dt} = 49.2 \, \text{A/s} \), we get:\( \varepsilon = -0.00369 \times 49.2 \approx -0.1815 \, \text{V} \).

Key Concepts

Solenoids

A solenoid is essentially a coil of wire that is tightly wound, usually in the shape of a cylinder. When an electric current passes through it, a magnetic field is created. Solenoids are important in the study of electromagnetism because they are foundational components in many electromagnetic devices. The strength of the magnetic field produced inside a solenoid depends on several factors:

• The number of coils or turns of the wire.
• The current flowing through the wire.
• The presence of a material core that can increase the magnetic field strength.

In the context of the exercise given, two solenoids are placed within one another. This arrangement allows for the potential study of mutual inductance, where changes in current in one solenoid can induce electromagnetic effects in the other.

Electromagnetic Induction

Electromagnetic induction is a principle where a change in magnetic field within a closed loop of wire induces a voltage, or electromotive force (emf), in the wire. This is a foundational concept in electromagnetism, vital for understanding how many electrical devices work.

• Faraday's Law of Induction states that the induced emf in any closed circuit is equal to the time rate of change of the magnetic flux through the circuit.
• This means if the magnetic field changes due to a changing current, the change will induce an emf in nearby wires or coils, like our solenoids.

In our exercise, the changing current in the outer solenoid changes the magnetic field. This change induces an emf in the inner solenoid according to the laws of electromagnetic induction.

EMF Calculation

The electromotive force (emf) is calculated using the principle of mutual inductance, whereby we determine how the change in current of one solenoid affects another. The formula we used to calculate the mutual inductance M is:\[ M = \frac{\Phi}{I} \]where \( \Phi \) is the magnetic flux due to the outer solenoid's field, and \( I \) is the current. In practice:

• The magnetic field created by the outer solenoid is dependent on its turns per unit length and current.
• The total magnetic flux \( \Phi \) is then dependent on this field, the area of the inner solenoid, and its number of coils.

The emf in the inner solenoid is then given by:\[ \varepsilon = -M \frac{dI}{dt} \]where \( \frac{dI}{dt} \) denotes the rate of change of current. This helps us determine the voltage induced in the solenoid due to changes in magnetic flux.