# [Magnetism] Tricky task on self-inductance of solenoid

• wetback
In summary, by using the inductance equation for a solenoid and the resistance equation for a conductor, the number of turns in the winding can be eliminated to calculate the self-inductance of the solenoid. This can be achieved by expressing the length of the wire and the radius of the solenoid in terms of other variables and then substituting them into the inductance equation. The result, 5.5*10-5 H, can be found by simple algebraic manipulation.
wetback

## Homework Statement

Problem:
A Solenoid has N turns of Copper wire (Cross-section area of the wire is 10-6 m2) The Length of the solenoid is 25*10-2 m and its Resistance is 0.5 Ω. What's the Self-Inductance of the solenoid?

## The Attempt at a Solution

How do you relate Resistance with Self-inductance to eliminate the need for (or find) the number of turns? All I could find by using R was the length of the whole wire(an unrolled solenoid). But this gives me nothing on the number of turns cause the diameter of the solenoid is not given.

Hi wetback! http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I reckon you should assume the turns in the winding are single-layer, tightly-wound and close-packed. This allows you to determine the diameter of the coil.

What formula will you then use to calculate inductance?

Last edited by a moderator:
Hey! As it often is, when you are desperate enough to look for help in a forum, you miraculously find the solution yourself! Eureka! And as always, it turns out to be pretty simple. Just some algebra.

If anyone ever stumbles upon this post looking for help, I'll explain what I did:

I used the inductance equation for a solenoid L=($\mu$N2A)/l
where $\mu$ - permeability constant(4$\pi$*10^-7), N - number of turns, A - the cross-section area of the solenoid cylinder shape and l - the length of the solenoid
I also used the resistance equation for a conductor(this case the copper wire): R=($\rho$*l1)/A1 where $\rho$-the resistivity of copper(1.7*10-8), l1 - the length of the wire, A1- the cross-section area of the wire.
I expressed l1: l1=R*A1/$\rho$, then expressed N using l1: N=l1/(2$\pi$*r) where r - the radius of the solenoid cylinder shape. Then r=l1/(2$\pi$N) Now I expressed A: A=$\pi$*r2. When you insert this expression in the L equation, N disappears and we are left with just the given numbers and L really is 5.5*10-5 H.

## 1. What is self-inductance and how does it relate to magnetism?

Self-inductance is the property of a conducting material or circuit that causes it to oppose any change in current flow through it. This is due to the magnetic field generated by the current, which induces an opposing current within the material. This relationship between current and magnetic field is known as Faraday's law of induction.

## 2. What is a solenoid and how does it exhibit self-inductance?

A solenoid is a coil of wire that is tightly wound in a helical shape. When a current flows through the solenoid, it creates a magnetic field that is concentrated within the coil. This magnetic field induces an opposing current within the solenoid, causing self-inductance to occur.

## 3. How is self-inductance of a solenoid calculated?

The self-inductance of a solenoid can be calculated using the formula L = μ₀n²A, where L is the self-inductance in henrys, μ₀ is the permeability of free space, n is the number of turns in the solenoid, and A is the cross-sectional area of the solenoid. This formula assumes that the solenoid is tightly wound and has a uniform magnetic field.

## 4. What are some real-world applications of self-inductance in solenoids?

Solenoids are commonly used in electromechanical devices such as relays, motors, and generators. These devices rely on the principle of self-inductance to function properly. Additionally, self-inductance plays a crucial role in the operation of transformers, which are used to step up or step down the voltage of alternating current.

## 5. How is self-inductance of a solenoid affected by its physical properties?

The self-inductance of a solenoid is directly proportional to the number of turns in the coil and the cross-sectional area of the solenoid. It is also affected by the material used for the core of the solenoid, as well as the presence of any magnetic materials nearby that can influence the magnetic field. The spacing between the turns of the coil can also affect the self-inductance, with tighter winding resulting in a stronger magnetic field and therefore a higher self-inductance.

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