4
$\begingroup$

Few days ago I came by a derivation of STR from Bio-Savart law. Since then I have been trying to derive Bio-Savart law from STR. The derivation mentioned previously used two parallel current-carrying wires of unit length and used the change of their current density to achieve Lorentz contraction (for the entire derivation, go to http://www.andrijar.com/sr/sr.htm). Using the same idea, I have been able to derive ampere force law(F=2kII'/r) from STR. So,

Is it possible to derive Biot-Savart law from ampere force law? If yes, how?

(N.B. Please don't use Maxwell's equations for I have not yet learnt them)

$\endgroup$
4
$\begingroup$

Consider a positive point charge moving in positive $z$ direction at a constant velocity, at that instant when it's at the origin. The magnetic-field lines generated in plane $z=a$ are closed loops, further we observe circular symmetry in the problem about $z$-axis, henceforth the loops are circles.

A image to develop a idea of how we would be going to proceed:

A image to develop a idea of how we would be going to proceed:

Let's apply Ampere's displacement law to calculate magnetic field. Intuitively we can think, as the particle moves towards the plane $z=a$, the electric flux passing through a surface bounded by any circle centered on the z-asix in that plane will increase.

So, starting with Ampere's Law we have:

$$\oint \vec{B}.d\vec{I} = \mu_o \epsilon_o \frac{d\phi}{dt}$$

We integrate the left hand-side around a circle of radius $b$ on $z$-axis in plane $z=a$, and this is our Amperian Loop. so, $$ \oint \vec{B}.d\vec{I} = 2 \pi bB$$

To calculate electric flux $\phi$ enclosed by the circle of radius $b$, we select a spherical surface, of surface area $A$, which is bounded by the circle, is symmetric about z-axis, and has a radius $r$. So it follows from the understanding, we have: $$ \phi = \int \vec{E}.d\vec{A} = EA$$, where $$E= \frac{q}{4 \pi \epsilon_o r^{2}}$$

The area $A$ can be obtained using spherical, polar coordinates: $$ A= r^{2} \int_0^2\pi d\phi \int_0^\theta \sin x dx = 2 \pi r^2(1-\cos \theta)$$ $$\cos \theta= \frac{z}{\sqrt{z^2+x^2}}$$

Using/combining the above three equations we can have: $$\phi = \frac{q}{2 \epsilon_o}(1-\cos \theta)$$

Differentiating with time gives, $$\frac{d\phi}{dt} = -\frac{q}{2 \epsilon_o} \frac{d \cos \theta}{dt}$$ where, $$\frac{d \cos \theta}{dt} = \frac{d \cos\theta}{dz} \frac{dz}{dt}$$

In the above equation, $z$ is the distance between the particle, which is moving in the positive $z$ direction, and the Amperian Loop, which is fixed in $z=a$ plane. Since $z$ is decreasing with time, $$\frac{dz}{dt} = -v$$ where v is the speed of the particle. Now,

$$\frac{d \cos \theta}{dz}=\frac{y^2}{r^3}$$, where $r=\sqrt{z^2+y^2}$

Combining the first two and last three equations, we arrive at our awaited result as, $$B= \frac{\mu_o}{4\pi} \frac{qv \sin \theta}{r^2}$$, which can be vectorially rewritten as, $$\vec{B} = \frac{\mu_o}{4\pi} \frac{q \vec{v} \times \vec{r}}{r^3}$$

Hence, we are done! :)

$\endgroup$
  • $\begingroup$ But I asked for the derivation from AMPERE FORCE LAW, not just AMPERE'S LAW. The former states that the force between 2 parallel wires per unit length will be F=2kII'/r where k is the is the magnetic force constant, r is the separation of the wires, and I, I' are the direct currents carried by the wires (refer to link) $\endgroup$ – Ayan Biswas Mar 14 '15 at 1:41
  • 1
    $\begingroup$ This is a nice derivation so +1 but what does this have to do with SR? $\endgroup$ – FenderLesPaul Mar 18 '15 at 14:41
  • $\begingroup$ @FenderLesPaul the OP has already derived Ampere's law from SR and I derived Biot - Savart Law from Ampere's law and hence we can imply deriving Biot - Savart from SR. $\endgroup$ – ritvik1512 Mar 18 '15 at 14:43
  • $\begingroup$ The expression for the electric flux is wrong. The electric field of a moving charge is not the same as the electric field of a stationary charge. $\endgroup$ – Ben Crowell Jan 13 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.