# Magnetic induction in Relativity

As we know magnetic phenomenon is a mere relativistic effect.My question is how to explain the magnetic induction in a relativistic manner?

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Magnetism isn't just a relativistic effect. It's primarily a quantum mechanical one. – TeeJay Oct 5 '13 at 7:25
It's a matter of point of view really. Electrostatic can also be viewed as a mere relativistic effect applied to magnetism...it depends where you preferences lie. People learning electrotechnology for example focus much more on the Laplace-Lorentz force that the Coulomb one because this is the important thing to rememeber when dealing with electric motors. – gatsu Oct 6 '13 at 11:31

By using Coulomb law, $$\mathbf F = q\mathbf E = q\frac{Q\mathbf r }{|\mathbf r |^{3}},$$ and relativistic transformation laws, $$\mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r )}{c^{2}} - \gamma \mathbf u t , \quad \frac{\mathbf F }{\gamma \left( 1 - \frac{(\mathbf u \cdot \mathbf v)}{c^{2}}\right)} = \mathbf F ' + \gamma \frac{\mathbf v ' \cdot \mathbf F '}{c^{2}}\mathbf u + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F ') }{c^{2}}, \quad \Gamma = \frac{\gamma - 1}{\frac{u^{2}}{c^{2}}},$$ where $\mathbf u$ can be interpreted as charge $Q$ speed, $\mathbf v$ can be interpreted as test charge speed,
for $t = 0$, you will obtain an expression
$$\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B] ,$$ where $$\mathbf E = \frac{Q\gamma \mathbf r}{\left( r^{2} + u^{2}\gamma^{2}\frac{(\mathbf u \cdot \mathbf r)^{2}}{c^{2}}\right)^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E].$$ As you can see according to the $\mathbf F$ and $\mathbf B$ expressions, the induction is relativistic kinematic effect which is connected with finite speed of interactions. By the other words, it can be described as delay of the electric field displacement in time.