# Tag Info

6

Magnetic fields never do work directly. This is because the magnetic force on any charged particle, $$\mathbf{F}=q\mathbf{v}\times \mathbf{B,}$$ is always orthogonal to the velocity, and therefore the power transferred, $\mathbf{F}\cdot\mathbf{v}$, is zero. On the other hand, this seems to contradict much of our intuition about how magnets behave. If you ...

5

The are various crystal forms that iron and steel can adopt, the common ones being ferritic, martensitic and austenitic. The ferritic and martensitic forms are ferromagnetic (or just magnetic in everyday terms) while the austenitic form is not. So it isn't simply that iron is magnetic and steel isn't, it is specifically austenitic iron and steel that isn't ...

5

The field lines in your drawing are not the trajectories of photons. The field lines show the direction of the force on a test magnetic dipole. The force, and therefore its direction, is mediated by virtual photons (or can be described that way) but those photons will travel in straight lines just like ordinary photons.

3

All the filings are resting on a surface where there is friction. If the force due to the magnet does not exceed this friction, the filings won't accelerate to the magnet (only the closest ones experiencing the largest forces do). Iron is also ferromagnetic, which means it can concentrate magnetic fields. So filings that are too far to be affected by the ...

3

Instead of thinking about one field changing in response to the change of the other, it is more correct to say that whenever the magnetic field is changing, so is the electric field, and vice versa. The way these fields change is governed by Maxwell's equations. This way, we do not arrive at the confusion OP has.

3

First of all, you need to know what $\vec{M}$ is, which is a problem! Secondly, you are doing a volume integral, but if $m$ is constant, then from the point of view of the microscopic currents that create the dipoles, only the boundary or surface currents actually contribute (if you draw a lot of directed square loops on a piece of paper, you will see that ...

2

In principle you could think of it as the center of the dipole. The solution given, is for the field far from the dipole (distance from the dipole to the point much bigger than the size of the dipole). Given that, the size of the dipole is rather irrelevant and it's considered as a point.

2

It doesn't matter. If you have a scalar $\alpha$ then $$\alpha (\vec{B} \times \vec{C}) = (\alpha \vec{B}) \times \vec{C} = \vec{B} \times (\alpha \vec{C}).$$ You can prove this simply by writing out the components of each of the three expressions and showing that no matter which order you do it in you will get the same results.

2

There are 2 types of magnetic fields coming from phone: DC, from smartphone circuitry AC, from chip transmission (for each chip, if they are different,e.g Wi-Fi, Bluetooth, GSM. You will see different radiation pattern) It's better to say that voltage induced due to emission of your smartphone, electronics emission is DC. When you put screw you disturb ...

1

I'm not sure what kind of steel you are using, but with stainless steel, many varieties are slightly magnetic. As I recall, when we were building a spectrometer that was very sensitive to magnetic fields, we were careful to use only "austentic stainless steel", which is not magnetic. I think that we paid extra for this feature :) ...

1

The "lines" you see when viewing iron filings around a magnet have more to do with the fact that they are tiny slivers of iron, and less to do with magnetic field lines as one normally talks about them. Also, over the length scale of one of these slivers, the magnetic field is largely constant, and a ferromagnet (or magnetic dipole) placed in a constant ...

1

A couple problems with your development: 1) (minor, now corrected) You lose a couple factors of $\mu_0$ when you substitute for $B$ in the expression for $u_B$. 2) (major) You lose a minus sign in calculating the radial force; by your expression, the force should incorrectly tend to compress the coil, not expand it. What's going on? When mechanical ...

1

I can give you a simplified picture: The external magnetic field induces in the atoms of a diamagnetic material a current, which produces a magnetic field in the opposite direction (Lenz's law). Because of this effect, the diamagnet is repelled away. The same effect takes place in an paramagnet, but here is another effect stronger: Due to the magnetic ...

1

As Emilio Pisanty asserts, the magnetic Lorentz force $\boldsymbol{F_m}= q \boldsymbol{v \times B }$ does not do work, since $\boldsymbol{F_m \cdot v} = 0$. I don't have access to Griffith, but I find Feynman's discussion (in v. II, ch 15) confusing, since he has magnetic forces doing work on the current while also asserting those same forces do no work. ...

1

Since the arrangement is symmetrical when you interchange N and S, I would say that the maximum strength will come when D2 and D3 are equal. Also, the smaller D1 becomes, the more the flux lines will become compressed. Also, a sheet of metal where you indicate would have almost the same effect as making the magnets wider, without an increase in mmf. So I ...

1

As you already know, the effective length is the distance between the poles of a magnet. It is obvious that the poles are not found at the extremities of the magnet. Thus, the effective length $l_{eff}$ is shorter than the geometric length. Now I hope that I remember this part right (didn't have time to double check in books), but the effective length is ...

1

The answer has to do with relativity. You are correct in saying that in the particle's rest frame its velocity is zero and so there's no magnetic force, but the particle should still move. Special Relativity tells us that the electric and magnetic fields are really two aspects of the same thing instead of being distinct entities. When we change our reference ...

1

I believe the term you're looking for is "Magnetostatic Energy". Magnetostatics is the field that studies static (constant) magnetic fields, much like electrostatics. For a uniform material the magnetostatic stored potential energy is: $$E_{\mathrm{ms}} = \frac{1}{2}\mu_0 \int_V \mathbf{M} \cdot \mathbf{H}_{\mathrm{ms}} d^3 r$$ You can find a full ...

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