How Is the Self Inductance of a Solenoid Calculated?

[diegocl02]

Homework Statement

demostrate that the self inductance of a selonoid (L) that have: length = "l"; radius= "a" ; number of turns per unit of length= "m" , is:

Homework Equations

L = pi*(m²)*(a²)*U_o*[ ( (l²)+(a²) )^1/2) - a ]

The Attempt at a Solution

It is suposse that the selfinductance is just
L= pi*(m²)*(a²)*U_o * l

but i don't understand how they get the expression [ ( (l²)+(a²) )^1/2) - a ]

please help!

 

[anathema]

The expression [ ( (l2)+(a2) )1/2) - a ] comes from the formula for the self-inductance of a solenoid, which is given by:

L = μ₀n²A/l

Where:
μ₀ = permeability of free space
n = number of turns per unit length (m)
A = cross-sectional area of the solenoid (πa²)
l = length of the solenoid

Substituting these values into the formula, we get:

L = μ₀(m²)(πa²)/l

Now, we need to simplify this expression to get it in terms of the given variables (l, a, and m). To do this, we use the Pythagorean theorem, which states that for a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. In this case, the hypotenuse is the length of the solenoid (l) and the other two sides are the radius (a) and the height (h) of the solenoid.

So, we can rewrite the expression for the cross-sectional area (A) as:

A = πa² = π(a² + h²)

Solving for h, we get:

h = (l² - a²)1/2

Substituting this into the original formula for self-inductance, we get:

L = μ₀(m²)(π(a² + (l² - a²))) / l

Simplifying this further, we get:

L = μ₀(m²)(π(l²)) / l

Finally, we can rewrite this in terms of the given variables as:

L = μ₀(m²)(π(l² + a²)) / l - μ₀(m²)(πa²) / l

Which can be further simplified to:

L = μ₀(m²)(π(l² + a²) - πa²) / l

And finally, we get:

L = μ₀(m²)(π(l²) - πa²) / l

Which is the same as [ ( (l2)+(a2) )1/2) - a ].

 

[kerimek]

The expression [ ( (l2)+(a2) )1/2) - a ] comes from the calculation of the magnetic flux through the solenoid. The self-inductance of a solenoid is a measure of the ability of the solenoid to generate an electromotive force (EMF) in itself when the current through it is changing.

The magnetic flux through a solenoid is given by the equation Φ = B*A, where B is the magnetic field and A is the cross-sectional area of the solenoid. For a solenoid with a circular cross-section, the area A can be approximated as A = π*a^2, where a is the radius of the solenoid.

The magnetic field inside the solenoid can be calculated using the equation B = μ0*n*I, where μ0 is the permeability of free space, n is the number of turns per unit of length (m), and I is the current through the solenoid.

Substituting these equations into the expression for magnetic flux, we get Φ = μ0*n*I*π*a^2. This is the total magnetic flux through the solenoid.

Now, for a changing current, the EMF induced in the solenoid is given by the equation EMF = -N*dΦ/dt, where N is the number of turns in the solenoid and dΦ/dt is the rate of change of magnetic flux. In this case, since the current is changing, the magnetic flux is also changing, and the EMF induced in the solenoid is given by EMF = -N*μ0*n*dI/dt*π*a^2.

Finally, the self-inductance of the solenoid is defined as L = EMF/dI, which gives us L = μ0*n*π*a^2. This is the expression that you mentioned in your attempt at a solution.

However, this expression does not take into account the length of the solenoid. The correct expression for the self-inductance of a solenoid is L = μ0*n*π*a^2*[ ( (l^2)+(a^2) )^(1/2) - a ]. This takes into account the length of the solenoid and gives a more accurate value for the self-inductance. So, the expression [ ( (l^2)+(