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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$\begingroup$ Magnetism isn't just a relativistic effect. It's primarily a quantum mechanical one. $\endgroup$– TeeJayCommented Oct 5, 2013 at 7:25
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$\begingroup$ 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. $\endgroup$– gatsuCommented Oct 6, 2013 at 11:31
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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.
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$\begingroup$ ,My question is how to understand the force acting on a charged particle due to magnetic flux variation without the need for magnetism?The same way relativity has shown that magnetism is nothing but a revelation of electric field plus relativistic length contraction effect.Is it possible? $\endgroup$ Commented Oct 5, 2013 at 18:29
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$\begingroup$ @MuhammadMorrsi . You may not use the term "magnetic field" name and work only with electric field, but it is senselessly. Did I understand your question correctly? $\endgroup$– user8817Commented Oct 5, 2013 at 23:41