# Self inductance of a solenoid

## The Attempt at a Solution

Self inductance L = μ0n2πr2l . This means it depends only on geometrical factors .

So , self inductance in both the cases should be equal .

But if I think in terms of L = Φ/I , then Magnetic field at the center is twice that at the ends (assuming long solenoid ) . In that case flux at the center will be twice that at the ends and so will be the self inductance .

I am not sure which of the above two reasoning is correct .

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Homework Helper
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The formula you have is for the self inductance of a very long solenoid. You are correct that the magnetic field is no longer uniform at the ends of a solenoid of finite length, and is approximately (almost exactly) 1/2 at the ends, and thereby the ## -\frac{d \Phi}{dt} ## will be one-half as much per unit length near the ends as anywhere near the middle. Since inductance ## L=\frac{\Phi}{I} ## the effect on the inductance in the region near the ends is that it contributes half as much per unit length. ## \\ ## Incidentally, one can take the deivative of the last equation and write that the EMF in the inductor is ## \mathcal{E}=-\frac{d \Phi}{dt}=-L \frac{dI}{dt} ##.

Jahnavi
Do you mean SI at the center is greater ?

You are correct that the magnetic field is no longer uniform at the ends of a solenoid of finite length, and is approximately (almost exactly) 1/2 at the ends,

I think this is for a long solenoid , not a finite length solenoid .

Homework Helper
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The magnetic field of a long solenoid is ## B_z=n \mu_o I ## throughout its entire length and is very uniform with ## B_r=B_{\phi}=0 ##. ## \\ ## It is a very well-known (or somewhat well-known) problem from advanced E&M theory that the magnetic field ## B_z ## at the openings of the solenoid is ## B_z=\frac{1}{2}n \mu_o I ##. ## \\ ## For the details: This result is because the magnetic field of a magnetized cylinder (with the same dimensions as the solenoid) with uniform magnetization ## M_z ## in the z-direction has the exact same magnetic field pattern as the solenoid, (both inside and outside the cylinder), with ## B_z= M_z ## inside and anywhere near the middle region. Some very straightforward calculations using the magnetic pole method give the result that there are surface magnetic charge densities ## \sigma_m=\vec{M} \cdot \hat{n} ## at the endfaces, resulting in a contribution of ## B_z=-(1/2) M_z ## inside the cylinder near this "plane of surface charge", (comes from ## H_z=-\frac{\sigma}{2 \mu_o} ## inside the cylinder, near the endface, with an inverse square law fall-off as one moves away from the endface), and ## \vec{B}=\mu_o \vec{H}+\vec{M} ##. (Note: This result is quite basic for understanding the concept of magnetic surface concepts in the magnetic surface current methodology for computing the magnetic field of a permanent magnet. Notice how the magnetic surface current per unit length ## \vec{K}_m=\vec{M} \times \hat{n}/\mu_o ## corresponds to the solenoid current per unit length of ## nI ##).## \\ ## In any case, this is a well-known result that is presented in advanced E&M courses, and so the result can be used without going through all of the details. And your second line of reasoning in the OP is the correct one.

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Jahnavi
OK .

Do you agree that , considering energy density = (1/2)B20 and , with the similar reasoning as in the OP , energy density at the center is greater ?

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OK .

View attachment 226658

Do you agree that , considering energy density = (1/2)B20 and , with the similar reasoning as in the OP , energy density at the center is greater ?
That is correct.

Jahnavi
Thanks .

Could you explain qualitatively what does self induction per unit length signify ? I mean what does it tell you about the solenoid ? Why per unit length ?