What is the self inductance of a tightly wound solenoid with a thick wire?

In summary, to calculate the self inductance of a solenoid, we can use the formula L=u0n2V where V is the volume of the solenoid, n is the number of turns per unit length, and u0 is a constant. If the wire is thin enough, we can simply use this formula. However, if the wire is not negligible, we can use the formula n=N/L to determine n, where N is the total number of turns and L is the length of the solenoid. Once we have determined n, we can plug it into the formula L=u0n2V to calculate the self inductance. In this case, the self inductance of the solen
  • #1
Abhishekdas
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Homework Statement


Calculate the self inductance of a solenoid that is tightly wound with a wire of diameter .1 cm , has a cross sectional area of .9 cm2 and is 40 cm long?


Homework Equations






The Attempt at a Solution



the formula L=u0n2V where V=volume and n=number of turns per unit length is valid when we assume the wire of solenoid is very thin. Now if the wire is so thick that it is not negligible hen how to find the self inductance?
 
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  • #2
Since the wire has a diameter of .1 cm, we can assume that it is thin enough to use the formula given above. Therefore the self inductance is equal to u0n2V, where V=cross sectional area x length = .9 cm2 x 40 cm = 36 cm3 and n=number of turns per unit length. To determine n, we can use the formula n=N/L where N is the total number of turns in the solenoid and L is the length of the solenoid, which in this case is 40 cm. Therefore, if the solenoid has a total of 400 turns, then n=400/40=10 turns/cm. Therefore, the self inductance of the solenoid is equal to u0n2V = (4pi x 10-7) x (10 turns/cm)2 x (36 cm3) = 4.32 x 10-3 H.
 

1. What is self inductance?

Self inductance is a property of an electrical circuit that describes the ability of the circuit to generate an electromotive force (EMF) in itself when the current through it changes.

2. How is self inductance calculated?

Self inductance can be calculated by taking the ratio of the magnetic flux generated by a current in the circuit to the current itself. It is also dependent on the geometry and material properties of the circuit.

3. How does self inductance affect a circuit?

Self inductance can affect a circuit by creating back EMF, which opposes changes in current flow. This can cause a delay in the circuit's response to changes in input and can also lead to energy losses.

4. What are some practical applications of self inductance?

Self inductance is used in many electrical devices, such as transformers, motors, and generators, to control and manipulate the flow of current. It is also used in power transmission to reduce energy losses.

5. How can self inductance be reduced?

Self inductance can be reduced by using materials with lower magnetic permeability, such as copper or aluminum, and by using designs that minimize the length and number of turns in the circuit. Additionally, using materials with higher electrical conductivity can also help reduce self inductance.

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