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Nandhitha1098

A long narrow solenoid has length l and a total of N turns, each of which has cross-sectional area A. Its inductance is μ₀ ( [tex]\frac{N^2A}{l}[/tex] ).

To find the answer, we need to know about the magnetic field due to a solenoid.

What is the magnetic field due to a solenoid?

  • A solenoid is an insulated wire wound in the form of a helix.
  • Its length is very large compared to its diameter.
  • The magnetic field due to solenoid carrying current [tex]I[/tex] can be written as,

                             [tex]B=[/tex] μ₀n [tex]I[/tex]

                               [tex]n=\frac{N}{l}[/tex]

  • We have an expression for magnetic flux as,

                            Φ [tex]= BA[/tex] Where, A is the area.

  • Substituting above value of magnetic field in the expression of magnetic flux, we get,

                                       Ф [tex]= \frac{NIA}{l}[/tex] μ₀

  • In terms of inductance, we can express the magnetic flux linked with the coil as,

                                        Ф [tex]= LI[/tex]

  • By equating above two equations of magnetic flux, we get the expression for inductance as,

                                        μ₀ ( [tex]\frac{N^2A}{l}[/tex] ) [tex]I[/tex]  [tex]=LI[/tex]

                                         [tex]L=[/tex] μ₀ ( [tex]\frac{N^2A}{l}[/tex] ).

Thus, we can conclude that the expression for inductance is  μ₀ ( [tex]\frac{N^2A}{l}[/tex] ).

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The inductance of a solenoid having N number of turns, area A and a length l will be equal to μ₀ ( [tex]\frac{N^2A}{l}[/tex]).

We need to understand the magnetic field created by a solenoid in order to determine the solution.

What does a solenoid's magnetic field look like?

  • An insulated wire twisted into a helix is what makes up a solenoid.
  • Compared to its diameter, its length is significantly larger.
  • It is possible to express the magnetic field caused by a solenoid carrying current as,

                                    [tex]B=\frac{NI}{l}[/tex]×μ₀

  • We can express magnetic flux as follows:

                                   [tex]flux=BA[/tex]  , A is the area.

  • In the formula for magnetic flux, we obtain by substituting the above magnetic field value:

                                      [tex]flux=\frac{NIA}{l}[/tex] times the μ₀.

  • The magnetic flux associated with the coil can be expressed in terms of inductance as follows:

                                        [tex]flux=IL[/tex]

  • We obtain the expression for inductance as, by combining the aforementioned two magnetic flux equations.

                                     μ₀ ( [tex]\frac{N^2AI}{l}[/tex])= [tex]LI[/tex]

                                     [tex]L=[/tex] μ₀ ( [tex]\frac{N^2A}{l}[/tex]).

As a result, we can say that the inductance expression is   [tex]L=[/tex] μ₀ ( [tex]\frac{N^2A}{l}[/tex]).

Learn more about solenoids here:

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