Question

One solenoid is centered inside another. The outer one has a length of 50.0 cm and contains 6750 coils, while the coaxial inner solenoid is 3.0 cm long and 0.120 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 inner solenoid.

Short answer

The mutual inductance is approximately 2.88 μH, and the induced emf is about -142 μV.

Step-by-step solution

Understanding the Problem

We need to find the mutual inductance ( M ) between two solenoids and the induced emf ( ε ) in the inner solenoid due to the changing current in the outer solenoid.

Setting Up the Formula for Mutual Inductance

The formula for the mutual inductance between two solenoids is given by:\[M = \mu_0 \frac{N_1 N_2 A}{l_1}\]where \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)is the permeability of free space, \(N_1\)and \(N_2\)are the number of turns in the outer and inner solenoids, \(A\)is the cross-sectional area of the inner solenoid, and \(l_1\)is the length of the outer solenoid.

Calculate the Cross-Sectional Area

The diameter of the inner solenoid is 0.120 cm.Convert it to meters: \(0.120 \, \text{cm} = 0.0012 \, \text{m}\).The cross-sectional area (A) is given by:\[A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0012}{2}\right)^2 \approx 1.13 \times 10^{-6} \, \text{m}^2\]

Substitute Values into Mutual Inductance Formula

Substitute the known values into the formula:\[M = \left(4\pi \times 10^{-7}\right) \times \frac{6750 \times 15 \times 1.13 \times 10^{-6}}{0.5}\]Calculate:\[M \approx 2.88 \times 10^{-6} \, \text{H}\]

Determine the Induced EMF in the Inner Solenoid

The formula for the induced emf (ε) in the inner solenoid is given by:\[ε = -M \frac{dI}{dt}\]where \(dI/dt\)is the rate of change of current, which is 49.2 A/s.

Substitute Values into EMF Formula

Substitute the known values into the emf formula:\[ε = -2.88 \times 10^{-6} \times 49.2\]Calculate:\[ε \approx -1.42 \times 10^{-4} \, \text{V} = -142 \, \text{µV}\]

Conclusion

Thus, the mutual inductance of the solenoids is \(M \approx 2.88 \times 10^{-6} \, \text{H}\), and the emf induced in the inner solenoid is approximately \(-142 \, \text{µV}\).

Key concepts

Solenoids

Solenoids are a type of electromagnet and consist of a wire coiled into a cylindrical shape. They have a wide range of applications, especially where controlled magnetic fields are essential.
Solenoids can be instrumental in many devices such as electric bells, valves, and switches. They work on the principle of turning electrical energy into mechanical work.

• The magnetic field inside a solenoid is strongest and highly uniform when the solenoid is long, cylindrical, and tightly wound.
• Combining solenoids can create interesting electromagnetic effects, such as the mutual inductance explored here.
• When two solenoids are positioned together, as in the problem, they can influence each other magnetically. Changes in the current of one solenoid can induce a voltage in the other.

Understanding solenoids' functioning and their interactions help solve problems involving mutual inductance effectively.

Induced EMF

Induced Electromotive Force (EMF) occurs when a conductor, like a solenoid, experiences a change in magnetic flux. In our case, the EMF is induced in the inner solenoid due to the changing current in the outer solenoid.
According to Faraday's Law of Induction, a changing magnetic field within a closed loop induces an EMF in the loop. This induced EMF is crucial in many technologies, including transformers and electric generators.

• The key to understanding induced EMF is the rate of change of current. Faster changes in current result in higher induced EMF.
• The formula: \[ε = -M \frac{dI}{dt}\] indicates that the direction and magnitude of the induced EMF are dependent on both the mutual inductance (M) and the rate of current change (\(dI/dt\)). The negative sign represents the direction of the induced EMF, as per Lenz's law, which ensures that the induced EMF opposes the change in magnetic flux.

In this problem, the induced EMF in the inner solenoid is calculated using these principles, emphasizing the role of mutual interaction between solenoids.

Permeability of Free Space

The permeability of free space, denoted as \( \mu_0 \), is a fundamental constant in physics. It describes the ability of a vacuum to support a magnetic field. Its value is \( 4\pi \times 10^{-7} \) T m/A.
This constant plays a vital role in the equations governing electromagnetism and proves essential when analyzing magnetic phenomena such as mutual inductance.
The concept of permeability helps us quantify how much magnetic field is "allowed" in a vacuum or in material. That is why permeability is vital for calculations involving magnetic fields, like those between solenoids.

• In the exercise, \( \mu_0 \) is used to determine the mutual inductance between the two solenoids through the formula: \[M = \mu_0 \frac{N_1 N_2 A}{l_1}\] This formula highlights the proportional relationship between mutual inductance and permeability.
• This constant also impacts the electrical properties of materials and their interaction with magnetic fields.

By understanding permeability and its role, one can better grasp the mechanisms behind solenoids and induced EMF.