0
$\begingroup$

Two long concentric cylindrical conductors of radii a and b (b<a) are maintained at a potential difference V and carry equal and opposite currents I. An electron with a particular velocity 'u' parallel to the axis will travel un-deviated in the evacuated region between the conductors. Then u=?

The solution key contains the term for potential, this I can't understand because I had thought that when we having moving charges the electric fields transforms into magnetic fields. But here, the solution suggests that I should that I should account for both electric and magnetic fields.

In a similar post mine to mine, one of the users gave an answer that existence of moving charge and field charge together at same time is compatible, but how? How can you have free movement of charge to produce a magnetic field but at the same time some fixed charged which produces Electric field?

$\endgroup$
1
2
$\begingroup$

As SuperfastJellyfish points out in the comment above, a moving charge creates both an Electric Field and a Magnetic Field (a very good reference for this is the Feynman Lectures on Physics).

You can't always make an electric field "disappear". There are many ways to see it, but I feel the simplest is to realise that the quantity (call it $Q$) $$Q = \mathbf{E}^2 - c^2 \mathbf{B}^2$$ is a Lorentz invariant, meaning in a new frame even if $\mathbf{E \to E'}$ and $\mathbf{B \to B'}$, the quantity $Q = \mathbf{E'}^2 - c^2 \mathbf{B'}^2$ will remain the same.

Consider a frame in which there is no magnetic field. (Say the rest frame of the charge in question.) Then $$Q = \mathbf{E}^2 > 0.$$ Now let's move to another frame in which there is an electric and a magnetic field $\mathbf{E'}$ and $\mathbf{B'}$ respectively. (This could be the lab frame in which the charge is moving with respect to us.) We must still have that $$Q = \mathbf{E'}^2 - c^2 \mathbf{B'}^2 > 0.$$

Can we ever have that $\mathbf{E'}=0$? No! Since that would mean that $-c^2 \mathbf{B'}^2 >0$, which is nonsense since the left hand side is a negative number.

$\endgroup$
0
$\begingroup$

it seems you need 3 different voltages, the first between the ends of the inner tube, the second at the outer tube, producing the currents in the tubes and a third connected between the inner and outer tube this third will produce an electric field between the cylinders.

$\endgroup$

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.