Self-Inductance of a Solenoid
Learning Goal:
To better understand self-inductance, using the example of a long solenoid.
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 n 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.
As the current through the solenoid changes, the resulting magnetic flux through the solenoid will also change, and an electromotive force will be generated across the solenoid according to Faraday’s law of induction:
E=???total?t,
where ?total is the total magnetic flux passing through the solenoid.
The self-inductance L is defined to be L=?total/I, where I is the current passing through the solenoid. Using the self-inductance, Faraday’s law can be rewritten as E=?L?I/?t.
The direction of the emf can be determined using Lenz’s law: The induced emf always opposes any change in the current I.
Part B
What is the magnetic flux ?1 through a single turn of the solenoid?
Express your answer in terms of the magnetic field B, quantities given in the introduction, and any needed constants.
?1 =
Part C
Suppose that the current varies with time, so that ?I/?t?0. Find the total electromotive force E induced in the solenoid due to this change in current.
Express your answer in terms of ?I/?t, n, Z, and R.
E =
Part D
The self-inductance L is related to the self-induced emf E by the equation E=?L?I/?t. 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.
L =
The direction of the induced emf can be determined using Lenz’s law. In the figure, the current I flows from A to B. Suppose that this current is decreasing with time; i.e.,?I/?t<0. The induced emf will tend to oppose this change, and must therefore be in the direction from A to B. End B will then be at a higher potential than end A. In other words, there will be a potential (or voltage) increase from A to B. The magnitude of this potential increase will be equal to L|?I/?t|.
Part E
Which of the following is always a true statement?
Which of the following is always a true statement?
The Correct Answer and Explanation is :
For the self-inductance problems you’re working on, here’s a breakdown to address each part step by step:
Part B: Magnetic Flux (\Phi_1) Through a Single Turn of the Solenoid
The magnetic flux (\Phi_1) through a single turn of a solenoid can be expressed by the formula for magnetic flux (\Phi = B \cdot A), where (B) is the magnetic field inside the solenoid and (A) is the cross-sectional area through which the field is passing.
The magnetic field (B) inside a long solenoid is given by (B = \mu_0 n I), where (\mu_0) is the permeability of free space, (n) is the number of turns per unit length, and (I) is the current. The cross-sectional area (A) of the solenoid is (\pi R^2) (where (R) is the radius of the solenoid).
Thus, the magnetic flux (\Phi_1) through one turn of the solenoid is:
[
\Phi_1 = B \cdot A = \mu_0 n I \cdot \pi R^2
]
Part C: Total Electromotive Force (E) Induced in the Solenoid
The total electromotive force (E) induced in the solenoid due to a change in current can be found by multiplying the rate of change of the magnetic flux by the total number of turns. From Faraday’s law of induction, (E = -N \frac{d\Phi}{dt}), where (N) is the total number of turns (in this case (N = nZ), where (Z) is the length of the solenoid).
Using the expression for (\Phi_1) from Part B and substituting:
[
E = -nZ \frac{d}{dt}(\mu_0 n I \pi R^2) = -\mu_0 \pi R^2 n^2 Z \frac{dI}{dt}
]
Part D: Self-Inductance (L) of a Long Solenoid
Self-inductance (L) is defined as (L = \frac{N \Phi_1}{I}). Substituting the values we get:
[
L = \frac{nZ \cdot \mu_0 n I \pi R^2}{I} = \mu_0 \pi R^2 n^2 Z
]
This represents the self-inductance in terms of (n), (R), (Z), and (\mu_0).
Part E: True Statement Regarding Induced EMF
From the analysis, the correct statement about the induced EMF, especially considering Lenz’s Law, would be that the induced EMF opposes the change in current. This means when the current decreases ((\frac{dI}{dt} < 0)), the induced EMF acts in a direction to oppose the reduction (trying to increase the current or at least oppose its decrease). Thus, if (I) flows from (A) to (B), and it is decreasing, the induced EMF will create a potential difference such that (B) is at a higher potential compared to (A), resulting in a potential increase from (A) to (B).
Therefore, the always true statement regarding the direction and effect of the induced EMF is:
- “The induced EMF due to a decreasing current increases the potential from the start to the end of the solenoid.”
This explanation adheres to the fundamental principles of electromagnetism, ensuring that Lenz’s Law, which dictates the direction of induced currents and electromotive forces, is satisfied, thus conserving energy within the system.