Formula for Magnetic Force on a Current-Carrying Wire

Question

There seems to be a pretty standard formula, that if a wire of length $\ell$ carrying current $I$ is immersed in a magnetic field $B$, then the magnitude of the magnetic force is $$F_B = I \ell B sin \theta$$, where the direction of $F_B$ is the direction of $\ell \times B$ (determined using right-hand rule).

But, these properties can all be summarized into the equation $$dB = \frac{\mu_0}{4\pi} \frac{Id \ell \times r}{r^2}$$

I understand the first equation, but don't get the second one. How would I even read this equation or apply it? Do I have to integrate first? How would this work?

Thank you in advance for any help/clarifications.

Answer 1

As the comments have already pointed out the two equations are related but not the same. The first equation represents the "magnetic force on a straight wire segment" due to a magnetic field $B$. That is to say, if you had a wire carrying current $I$ and "immersed it", as you put it, into a magnetic field of strength $B$, the force experienced by the wire is

$\vec{F} = I \vec{l} \times \vec{B} $

and the magnitude is given by $I l B sin \theta$, and the $sin \theta$ just comes from the cross product.

As you know $I$ is scalar but this force is a cross product so you use the direction the current is flowing in, denoted by $\vec{l}$, to figure out the direction. If needed I can provide the derivation of this equation, but I'd just be copy/pasting out of my book University Physics by Young and Freedman, Chp 27 section 27.6.

Now, if you are given a nice constant magnetic field, you can insert it into this equation and be done. But if you need to derive what the magnetic field is, due to some current carrying wire(s), solenoid, etc. you need to use Biot-Savart or Ampere's Law.

In general you get this from computing:

$\vec{B} = \frac{\mu_0}{4 \pi} \int \frac{I d\vec{l} \times \vec{r}}{r^2}$

So to get the force on a current carrying wire, from say another current carrying wire, you'd really need to do this:

$\vec{F} = I \vec{l} \times \frac{\mu_0}{4 \pi} \int \frac{I d\vec{l} \times \vec{r}}{r^2}$

Answer 2

"How would I even read this equation or apply it? Do I have to integrate first? How would this work?"

Using vector product notation, the equation (the Biot-Savart rule) may be written as

$$d\vec B=\frac{\mu_0}{4\pi}\frac{Id\vec l\times \hat r}{r^2}\ \ \ \text{that is} \ \ \ \ d\vec B=\frac{\mu_0}{4\pi}\frac{Id\vec l\times \vec r}{r^3}$$ in which $\vec{r}$ is the displacement vector from the 'current element' $Id\vec l$ to the point at which we want to know the field $d\vec B$ due to the current element. We write $|\vec r|$ simply as $r$, and the unit vector, $\frac{\vec r}r$, in the direction of $\vec r$ as $\hat r$.

So the first way of writing the B-S rule shows immediately that it's an inverse square law; the second avoids the use of the unit displacement vector, which some may regard as an unnecessary extra bit of notation.

In general to use this formula does require integration, but there's at least one important case where the integration is utterly trivial: finding the magnetic field at the centre (O) of a current-carrying circular loop... Since all current elements are the same distance, $r$, from O and at right angles to their displacements from O, we find $$\vec B=\frac{\mu_0}{4\pi}\frac{I\ 2\pi r}{r^2}\hat z\ \ =\ \frac{\mu_0 I}{2 r}\hat z$$ in which $\hat z$ is the unit vector parallel to the axis of the loop according to the right hand rule.