Recent content by Eric Wright

we have \vec E = - \nabla V - \frac{\partial \vec A}{\partial t} and the vector potential A can change due to the changing current thus the E field can be non zero

Take an arbitrary element of <2 \pi > and show it is in the kernel. Next take an arbitrary element of the kernel and show it is in <2 \pi > This is a technique called double containment and it is how we normally show that sets are equal.

1=cis(0) \in T the multiplicative identity in T is the number 1 which is e^0=cos0+isin0=cis(0) 1 is in T since 1=cis(0) and 0 is a real number

We are looking for the induced electric field.. not the B field. We can certainly find that there is some induced E field. We know the flux through a given cross section of the toroid is changing (since the current is changing) and thus there will be a circulation of electric field around a...

Look again... there is no horizontal strip on the bar unaffected by a slit. When he turns the bar you see there are extra slits to make sure of this. And the footage is sped up. The magnet in the tube with no slits takes 20 seconds to fall. The other magnet follows shortly after... surely much...

For #3 define: Note that R here is treated as an additive group! f: R \rightarrow T by x \rightarrow cis(x) problem 1 shows its an additive homomorphism you need to show that the kernel of f is < 2 \pi > do you know how to do this?

for #2 you need to verify the group axioms. pick z,w,v \in T you need to show: 1. zw \in T by the first problem you have proved... we have zw=cis(x)cis(y)=cis(x+y) \in T cis(x+y) is in T by definition since x+y is in R 2. (zw)v=z(wv) not hard to show... (since T is a subset of...

Actually I just found this video showing the magnet fall slowly in the tube with slits but not as slowly as compared to a tube without slits

I am hearing a lot of different conclusions here. Rude man says the magnet will fall as if there is no copper tube at all. Gordianus says the magnet will drop quicker when the slits are introduced but still be slowed a bit by the tube. What am I to make of this? Both of you seem to think the...

This doesn't seem right to me.. If we are viewing the tube as single-turn coils piled on top of each other, then when we cut slits in the coils, we have exactly the situation of the fat wire problem (cut a slit in a fat wire and it acts as a capacitor). The falling magnet certainly produces an...

That is what I was thinking at first.. but am not sure it is right. Not sure how to actually show that the field vanishes there. Yes, that is correct. sorry.. i should have been more clear

Homework Statement By considering the E and B fields of an incident monochromatic plane wave on a metal surface, as well as the current density of mobile electrons, J and the resulting EM force F felt by them, show that there is a force on the metal due to the magnetic force on the mobile...

I don't know. That is why I put together a model that views the coil as a circuit with capacitors. This led me to the conclusion that there will be less current in each coil as compared to a coil without slits, and thus less force. Is this a valid way to see the problem? What exactly do you...

Why not? There is a common problem where you view a slit in a fat wire as a capacitor. Why can't we use the same idea here? It leads me to the same conclusion that the magnet drops in less time if slits are added...

Homework Statement Suppose we have a toroidal solenoid with square cross section and whose axis of symmetry lies along the z axis. Suppose a current I(t)=kt runs through the solenoid. Find the induced electric field at an arbitrary point on the z axis. Homework Equations \oint \vec E \cdot...