Self-Inductance of a Solenoid Part A: Within the solenoid, but far from its ends, what is the magnetic field due to the current I? Express your answer in terms of some or all of the following variables: μ0, n, R, and any relevant constants. Learning Goal: To understand self-inductance, it is helpful to consider the specific example of a long solenoid, as shown in the figure. (Figure 1) This solenoid has radius R and length Z along the z-axis, and is wound with turns per unit length, so that the total number of turns is equal to nZ. Assume that the length of the solenoid is much greater than its radius. Part B: What is the magnetic flux through a single turn of the solenoid, Φ? The current passing through the solenoid is I. Using Faraday’s law and the concept of self-inductance, Faraday’s law can be rewritten as Φ = LI. The direction of the emf can be determined using Lenz’s law: The induced emf always opposes any change in the current I. Part C: Express your answer in terms of T/t, n, Z, and R. Part D: The self-inductance L is related to the self-induced emf Φ by the equation Φ = LI. Find L for a long solenoid. (Hint: The self-inductance L will always be a positive quantity.) Express the self-inductance in terms of the number of turns per length n, the physical dimensions R and Z, and relevant constants. Air-Core Solenoid Title_with_topic: Self-Inductance of a Solenoid and its Magnetic Field
The Correct Answer and Explanation is:
Part A: Magnetic Field Due to Current in the Solenoid
The magnetic field inside a long solenoid, but far from its ends, is given by Ampère’s Law. In this case, the magnetic field is uniform inside the solenoid, and the field strength depends on the current and the number of turns per unit length.
The magnetic field BBB inside the solenoid is given by:B=μ0nIB = \mu_0 n IB=μ0nI
where:
- μ0\mu_0μ0 is the permeability of free space (4π×10−7 T⋅m/A4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A}4π×10−7T⋅m/A),
- nnn is the number of turns per unit length,
- III is the current passing through the solenoid.
Part B: Magnetic Flux Through a Single Turn
The magnetic flux Φ\PhiΦ through a single turn of the solenoid is the product of the magnetic field BBB and the area AAA of the coil (the cross-sectional area of the solenoid):Φ=B⋅A\Phi = B \cdot AΦ=B⋅A
For a solenoid with radius RRR, the area of a single turn is:A=πR2A = \pi R^2A=πR2
Substituting the expression for the magnetic field BBB from Part A, we get the magnetic flux:Φ=μ0nI⋅πR2\Phi = \mu_0 n I \cdot \pi R^2Φ=μ0nI⋅πR2
Part C: Expressing Flux in Terms of T/t, n, Z, and R
To express the flux in terms of the time rate of change of the magnetic field (denoted as Tt\frac{T}{t}tT), we use Faraday’s Law of Induction:Induced emf=−dΦdt\text{Induced emf} = – \frac{d\Phi}{dt}Induced emf=−dtdΦ
By differentiating the flux Φ\PhiΦ with respect to time, we get:dΦdt=μ0nπR2dIdt\frac{d\Phi}{dt} = \mu_0 n \pi R^2 \frac{dI}{dt}dtdΦ=μ0nπR2dtdI
Thus, the induced emf is:emf=−μ0nπR2dIdt\text{emf} = – \mu_0 n \pi R^2 \frac{dI}{dt}emf=−μ0nπR2dtdI
Since dIdt\frac{dI}{dt}dtdI is the time rate of change of the current, we can rewrite it as Tt\frac{T}{t}tT, where TTT is the change in current over time ttt:emf=−μ0nπR2Tt\text{emf} = – \mu_0 n \pi R^2 \frac{T}{t}emf=−μ0nπR2tT
Part D: Self-Inductance of the Solenoid
The self-inductance LLL of the solenoid is defined by the equation:Φ=LI\Phi = L IΦ=LI
The magnetic flux Φ\PhiΦ through a single turn is μ0nπR2I\mu_0 n \pi R^2 Iμ0nπR2I, as shown earlier. Thus, using the definition of self-inductance:μ0nπR2I=LI\mu_0 n \pi R^2 I = L Iμ0nπR2I=LI
Solving for LLL:L=μ0nπR2ZL = \mu_0 n \pi R^2 ZL=μ0nπR2Z
where:
- ZZZ is the length of the solenoid.
Conclusion
- Part A: The magnetic field inside a long solenoid is B=μ0nIB = \mu_0 n IB=μ0nI.
- Part B: The magnetic flux through a single turn of the solenoid is Φ=μ0nIπR2\Phi = \mu_0 n I \pi R^2Φ=μ0nIπR2.
- Part C: The induced emf is emf=−μ0nπR2Tt\text{emf} = – \mu_0 n \pi R^2 \frac{T}{t}emf=−μ0nπR2tT.
- Part D: The self-inductance of the solenoid is L=μ0nπR2ZL = \mu_0 n \pi R^2 ZL=μ0nπR2Z.
This approach uses Ampère’s Law and Faraday’s Law to find the key properties of the solenoid related to its self-inductance.
