Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I do not know how to calculate the direction, or unit vector of the force that appear between two magnetic field sources. For example, let's assume I mean a current-carrying-wire by 'source of magnetic field'.

Image

In the image above, I know there is no force on circular wire, but why?

I want to know if we can only calculate the direction using a function of current and magnetic field vectors $\vec{F}(\vec{i},\vec{B})$, or is there a way to calculate the direction as a function of magnetic field vectors those are generated by the wires, $\vec{F}(\vec{B}_1, \vec{B}_2)$?

Do we take into account of the unit vectors on the intersecting points of magnetic fields those are generated by the wires?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Magnetic fields are generated by moving charged particles, and exert forces on moving charged particles.

The magnetic field generated by a moving charged particle can be calculated using Biot-Savart's law. Since a current is a steady stream of charged particles, the magnetic field created by a current-carrying wire can be calculated integrating over the length of the wire. The exact formula in classical electromagnetism is:

$$\mathbf{B} =\frac{\mu_0 q \mathbf{v}}{4\pi} \times \frac{\mathbf{r}}{r^2}$$

You want to pay close attention to the two vector quantities, v and r, the velocity of the charged particle and the position in space from it, and to the fact that they are combined with a cross product, $\times$. The resulting magnetic field, B, is therefore perpendicular to both. So a single electron moving in a straight line basically generates a magnetic field that goes around in circles around the path it is moving along. This picture may help.

Then you have the force acting on a moving particle when there is a magnetic field. This is known as Lorentz's Force, and the equation describing it is:

$$\mathbf{F} = q\ \mathbf{v} \times \mathbf{B}$$

You again have a cross product, this time involving the magnetic field, and again v, although this is a different v than before, not the velocity of the moving charged particle generating the field, but of the moving charged particle the field is acting on. Because of the cross product, a particle will not have any force acting upon it if it is moving parallel to the magnetic field. That's exactly why there is no force on the closed loop in your picture: because it lays along the direction of the magnetic field.

You could combine both equations into a single one, and so the force on particle 2 by particle 1, $\mathbf{F_{21}}$, if $\mathbf{r_{21}}$ is the position vector of particle 2 from particle 1, would then be

$$\mathbf{F_{21}} =\frac{\mu_0 q_1 q_2}{4\pi\ r^2}\mathbf{v_2} \times \mathbf{v_1} \times \mathbf{r_{21}}$$

share|improve this answer
    
thanks very much, the last equation is very explanatory. –  mervellous Aug 28 '12 at 8:09
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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