Suppose there is a test isolated charge and I am moving around it so in my frame of reference the charge is moving. So in my frame of reference does the charge produce a magnetic field? If yes, how can we calculate it? (which formula to use?)
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In lab frame, the charge is fixed, so it produces an electric field E. In the frame of the observer (consider inertial frame), charge is moving uniformly, then the observer will detect both electric and magnetic field. See this article https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#:~:text=Transformation%20of%20the%20fields%20between%20inertial%20frames,-The%20E%20and&text=Lorentz%20boost%20of%20an%20electric%20charge.&text=An%20observer%20in%20another%20frame,the%20motion%20of%20the%20charge. – KP99 Jul 18 '21 at 15:00
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In my answer here Electric field associated with moving charge the electric $:\mathbf{E}:$ and magnetic $:\mathbf{B}:$ parts of the electromagnetic field produced by a moving charge $:q:$ are given by equations (01.1) and (01.2) respectively being relativistic since they produced from the Liénard–Wiechert potentials...(1) – Frobenius Jul 18 '21 at 20:10
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(1)... Especially for a charge $:q:$ moving uniformly on a straight line the electric $:\mathbf{E}:$ and magnetic $:\mathbf{B}:$ parts of the electromagnetic field are given by equations (01a) and (01b) respectively in my answer here Magnetic field due to a single moving charge. – Frobenius Jul 18 '21 at 20:17
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Yes, the charge will produce a magnetic field in your frame of reference. The calculation of it can be difficult depending on the way you move around it. Most generally, the magnetic field in your frame can be calculated by taking the curl of the magnetic vector potential, which can be extracted from the Liénard–Wiechert potential which describes the effect of an arbitrarily moving electric point charge.
If your motion is uniform, however, it would be easier to simply calculate the electric field in the rest frame of the charge (the magnetic field being zero in this frame) and Lorentz transform into your rest frame to find the electric and magnetic fields in your frame.
Radu Moga
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Actually I am just in high school so I have not been taught how to convert the fields using Lorrentz transformastion. Although I have been taught about the Lorrentz force F=qE + q(VB) and the Biot Savart's law and Amperes circuital law. Also we have been restricted only to constant velocity so I just need to know about constant velocity. In this situation I can use the Biot-Savart's law which B=muqvsinx/4(pi)r^2,Right? – Harshal Chaware Jul 18 '21 at 15:27
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Yes, in your case of constant velocity (uniform motion) you can use Biot-Savart's law as you can imagine the charge moving on a straight line as a current with $I=q v$. Note, however, that in this case, $q$ is a line charge density. – Radu Moga Jul 18 '21 at 15:40
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"So in my frame of reference does the charge produce a magnetic field?"
No.
Charge doesn't not produce a magnetic field. The magnetic field is sourced from currents and changing currents, as seen on the past light cone (see https://en.wikipedia.org/wiki/Jefimenko%27s_equations).
However: one frame's charge is another frame's current:
$$\vec j = q\vec v$$
so that any frame that sees a moving charge will see a magnetic field.
The calculation of the fields is tractable, but tedious. A complete description is given in the venerable Feynman Lectures: https://www.feynmanlectures.caltech.edu/II_26.html
JEB
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Oh so basically the moving charge in my frame is nothing but a current for me and that is why it will produce a magnetic field, right? – Harshal Chaware Jul 19 '21 at 00:09
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@HarshalChaware A moving charge is not "nothing but at current" because it is not neutral, so it is both a charge and a current, and thus: sources both and electric and magnetic field. The standard current in a wire is neutral, so it only produces a magnetic field in the rest-frame of the ideal wire. (In other frames, it is not neutral, so it produces an additional electric field). – JEB Jul 24 '21 at 02:14