High pass filter - Cut-off frequency calculation doesn't match simulation

From the simulation, it looks like it fc=500Hz or somewhere around there.

Your simulation looks more like 15 kHz (3 dB point) and the reason is because you haven't used enough resolution in your AC analysis and the graph is inaccurately interpolating between points that are too far apart: -

Use more resolution in your AC analysis like in this simulation I did: -

The 3 dB point is around 7900 Hz.

Also try using a logarithmic X axis.

Well, mathematically speaking we can write:

$$\mathcal{H}\left(\text{s}\right)=\frac{\text{sL}}{\text{sL}+\text{R}}\tag1$$

Using \$\text{s}=\text{j}\omega\$, we get:

$$\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=\left|\frac{\text{j}\omega\text{L}}{\text{j}\omega\text{L}+\text{R}}\right|=\frac{\left|\text{j}\omega\text{L}\right|}{\left|\text{j}\omega\text{L}+\text{R}\right|}=\frac{\omega\text{L}}{\sqrt{\text{R}^2+\left(\omega\text{L}\right)^2}}\tag2$$

Now, we get:

• $$\lim_{\omega\to0}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=0\tag3$$
• $$\lim_{\omega\to\infty}\left|\underline{\mathcal{H}}\left(\text{j}\omega\right)\right|=1\tag4$$

Solving for the \$-3\space\text{dB}\$ point gives:

$$\frac{\omega\cdot10\cdot10^{-3}}{\sqrt{500^2+\left(\omega\cdot10\cdot10^{-3}\right)^2}}=\frac{1}{\sqrt{2}}\space\Longleftrightarrow\space$$ $$\omega=50000\space\text{rad/sec}\tag5$$

Which is the same as:

$$\frac{25000}{\pi}\space\text{Hz}\approx7957.75\space\text{Hz}\tag6$$