- #1

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- 17

## Main Question or Discussion Point

Consider two identical charges separated by some distance. Both are moving together with the same velocity, and the velocity vectors are perpendicular to a straight line drawn between the charges.

Each charge has charge

In the rest frame of both charges, the E-field at either charge is ##\frac{1}{4π ε_0}\frac{q}{r^2} = E_0##

Based on the Biot-Savart Law, the B-field at either charge is ##\frac{μ_0}{4π}\frac{qv}{r^2} = B_1##

##\frac{μ_0}{4π}\frac{qv}{r^2} = [v ⋅μ_0⋅ε_0] ⋅ [\frac{1}{4πε_0}\frac{q}{r^2}]##

##B_1 = \frac{v}{c^2}⋅ E_0##

Let's switch to the relativistic transformation for E and B fields: ##B_y'=γ(B_y+\frac{v}{c^2}⋅E_z)##

There is no magnetic field in the rest frame of both charges, hence ##B_y = 0##

Then ##B_y'=γ⋅\frac{v}{c^2}⋅E_z##

##E_z## is the E-field in the rest frame of both charges, hence ##E_z = E_0##

We thus obtain ##B_y'=γ⋅\frac{v}{c^2}⋅E_0##

We would expect ##B_y'=B_1## , but that is clearly not the case.

What is the explanation for the discrepancy?

Each charge has charge

**q**, speed**v**and they are separated by a distance of**r**.In the rest frame of both charges, the E-field at either charge is ##\frac{1}{4π ε_0}\frac{q}{r^2} = E_0##

Based on the Biot-Savart Law, the B-field at either charge is ##\frac{μ_0}{4π}\frac{qv}{r^2} = B_1##

##\frac{μ_0}{4π}\frac{qv}{r^2} = [v ⋅μ_0⋅ε_0] ⋅ [\frac{1}{4πε_0}\frac{q}{r^2}]##

##B_1 = \frac{v}{c^2}⋅ E_0##

Let's switch to the relativistic transformation for E and B fields: ##B_y'=γ(B_y+\frac{v}{c^2}⋅E_z)##

There is no magnetic field in the rest frame of both charges, hence ##B_y = 0##

Then ##B_y'=γ⋅\frac{v}{c^2}⋅E_z##

##E_z## is the E-field in the rest frame of both charges, hence ##E_z = E_0##

We thus obtain ##B_y'=γ⋅\frac{v}{c^2}⋅E_0##

We would expect ##B_y'=B_1## , but that is clearly not the case.

What is the explanation for the discrepancy?