- #1
Ahmes
- 78
- 1
The magnetic field of a moving charge is:
[tex]\boldsymbol{B} = \frac{\mu_0}{4\pi} \frac{q \boldsymbol{v}\times \boldsymbol{\hat{r}}}{r^2}[/tex]
This is an inverse square law.
But also we know that every localized current distribution (and a moving particle is most obviously a localized current distribution) appears from very far away as a dipole moment - which field is an inverse cube law.
Also using [itex]\boldsymbol{m} = \iiint \boldsymbol{x} \times \boldsymbol{J}(\boldsymbol{x}) d^3 x[/itex] it appears a moving charge, [itex]\boldsymbol{J}=q \boldsymbol{v} \delta^3 (x)[/itex] has a zero dipole moment.
So how could this be explained?
Thank you.
[tex]\boldsymbol{B} = \frac{\mu_0}{4\pi} \frac{q \boldsymbol{v}\times \boldsymbol{\hat{r}}}{r^2}[/tex]
This is an inverse square law.
But also we know that every localized current distribution (and a moving particle is most obviously a localized current distribution) appears from very far away as a dipole moment - which field is an inverse cube law.
Also using [itex]\boldsymbol{m} = \iiint \boldsymbol{x} \times \boldsymbol{J}(\boldsymbol{x}) d^3 x[/itex] it appears a moving charge, [itex]\boldsymbol{J}=q \boldsymbol{v} \delta^3 (x)[/itex] has a zero dipole moment.
So how could this be explained?
Thank you.
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