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Consider a unit charge in space it’s in rest with respect to first observer then the magnetic field produced by this according first observer is zero according to formula B=qvb let another observer is moving with velocity v and charge appears to be moving with respect to second observer so charge produces magnetic field according to second observer according to the formula B=qvb. How two different observers have different magnetic field for the same charge?

Sai Charan Reddy
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First, the expression $qvb$ is not the formula for the magnetic field of a moving electric charge. The correct expression is $F = qvB$, where $F$ is the force felt by an electric charge $q$ that is moving with velocity $v$ through a magnetic field $B$.

Second, it is true that different observers moving at different velocities won't agree on the strength of a magnetic field. The observer at rest with respect to the charge will measure zero magnetic field while the second observer will measure a non-zero field since the charge is moving in that reference frame. In special relativity, what looks like an electric field in one reference frame can look like a magnetic field in another, and like a combination of the two in a third frame.

Mark H
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    "In special relativity, what looks like an electric field in one reference frame can look like a magnetic field in another" - Can you fully cancel the electric field by a choice of a frame? – safesphere May 04 '18 at 05:53
  • @safesphere no you cannot since the “strength” of the electric charge - the source of the $\vec E$-field, is a Lorentz invariant. – ZeroTheHero May 04 '18 at 12:26